Electrochemistry Test

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Electrochemistry Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

Calculate the potential of an indicator electrode, which originally contained $$ 0.1 M \ MnO_4^- $$ and $$ 1.72 M \ H^+ , $$ and was treated with $$Fe^{2+} , $$ necessary to reduce 90% of $$ KMnO_4 $$ to $$Mn^{2+}$$.

Given: $$[E^0_{MnO_4^-} /_{ Mn^{+2} } = 1.51 V] $$

A
$$ 1.4 V $$

B
$$ 1.5 V $$

C
$$ 1.6 V $$

D
$$ 1.3 V $$

Solution
Question 2
1 / -0

What will be $$\Delta H$$ for the reaction; $$Ag(s)+\dfrac{1}{2}Hg_2Cl_2(s)\rightarrow AgCl(s)+Hg(l)$$ at $$25^0c$$, if this reaction can be conducted in a cell for which the emf = 0.0455 volt at this temperature with temperature coefficient $$3.389\times10^{-4} volt\, deg^{-1}$$?

A
+1280 cal

B
+640 cal

C
-1280 cal

D
-640 cal

Solution

$$Ag\left( s \right) +1/2{ Hg }_{ 2 }{ Cl }_{ 2 }\left( s \right) \longrightarrow AgCl\left( s \right) +Hg\left( l \right) $$
$$E=0.0455$$ volt $$\dfrac { dE }{ dT } =3.389\times { 10 }^{ -4 }$$Volt $${ deg }^{ -1 }$$
We know that, $$-nFE=\Delta H-nFT\left( \dfrac { dE }{ dT } \right) $$
$$\Rightarrow \quad E=-\dfrac { \Delta H }{ nF } +T\left( \dfrac { dE }{ dT } \right) $$
$$\Rightarrow \quad 0.0455=-\dfrac { \Delta H }{ 1\times 96485 } +298\times 3.389\times { 10 }^{ -4 }$$
$$\Rightarrow \quad \Delta H=+5354.16J=+1280Cal$$
Question 3
1 / -0

The density of copper is $$ 8 gm/cc.$$ Number of coulumbs required to plate an area of $$ 10 cm \times 10 cm $$ on both sides to a thickness of $$10^{-2} cm $$ using $$ CuSO_4$$ solution as electrolyte is:

A
$$24,250$$

B
$$48,124$$

C
$$96,500$$

D
$$10,000$$
Question 4
1 / -0

Consider the following half-cell reactions and associated standard half-cell potentials and determine the maximum voltage that can be obtained by combination resulting in spontaneous processes:
$$AuBr_{4}^{-}(aq) + 3e^{-} \rightarrow Au(s) + 4Br^{-}(aq); E^{\circ} = -0.86\ V$$
$$Eu^{3+}(aq) + e^{-} \rightarrow Eu^{2+}(aq); E^{\circ} = -0.43\ V$$
$$Sn^{2+}(aq) + 2e^{-} \rightarrow Sn(s); E^{\circ} = -0.14\ V$$
$$IO^{-}(aq) + H_{2}O(l) + 2e^{-}\rightarrow I^{-}(aq) + 2OH^{-}(aq); E^{\circ} = + 0.49\ V$$

A
$$+ 0.72$$

B
$$+ 1.54$$

C
$$+1.00$$

D
$$+ 1.35$$

Solution

$$\begin{array}{l} AuBr_{ 4 }^{ - }\left( { aq } \right) +3{ e^{ - } }\to Au\left( s \right) +4B{ r^{ - } }\left( { aq } \right) ;\, \, E^{ \circ }=-0.86V\to Anode \\ I{ O^{ - } }\left( { ar } \right) +{ H_{ 2 } }O\left( l \right) +2{ e^{ - } }\to { I^{ - } }\left( { aq } \right) +2O{ H^{ - } }\left( { aq } \right) ;E^{ \circ }=+0.49\to cathode \\ E{ ^{ \circ }_{ cell } }=E{ ^{ \circ }_{ cathode } }-E{ ^{ \circ }_{ anode } }\Rightarrow 0.49-\left( { -0.86 } \right) \Rightarrow +1.35V \end{array}$$
Question 5
1 / -0

$$E_{Fe^{+3}/ Fe^{+2}}^{0} = +0.77\ V; E_{Fe^{+3}/ Fe}^{0} = -0.036\ V$$. What is the value of $$E_{Fe/ Fe^{+2}}^{0}$$ ?

A
$$+0.44\ V$$

B
$$-0.44\ V$$

C
$$+0.48\ V$$

D
$$-0.48\ V$$

Solution
Question 6
1 / -0

Consider the half-cell reduction reactions :
$${ Mn }^{ 2+ }+{ 2e }^{ - }\rightarrow Mn,{ E }^{ o }=-1.18\ V$$
$${ Mn }^{ 2+ }\rightarrow { 2n }^{ 3+ }+{ e }^{ - },{ E }^{ o }=-1.51\ V$$
The $${ E }^{ o }$$ for the reaction $${ 3Mn }^{ 2+ }\rightarrow { Mn }^{ 0 }+{ 2Mn }^{ 3+ }$$ and possibility of the forward reaction are respectively:

A
-4.18 V and yes

B
+0.33 V and yes

C
+2.69 V and no

D
-2.69 V and no

Solution

The standard electrode potential of reaction can be calculated as,

$$E_{ 0 }=E_{ R }−E_{ P }$$

Where, $$E_{ R }$$ is the SRP of the reactant, $$E_{ P }$$is the SRP of the product.

$${ Mn }^{ 3+ }→Mn^{ 2+ }$$this reaction takes place with the $$E_{ 0 }-1.51 V$$

$$Mn^{ 2+ }→Mn$$ this takes place with$$E_{ 0 } -1.18 V$$

$$E_{ 0 }=−1.51 V−1.81 V$$

$$=-2.69V$$

and the reaction is will not occurs due to spontaneity.
Question 7
1 / -0

A current of $$i$$ ampere was passed for $$t$$ sec, through three calls $$P,Q$$ and $$T$$ connected in series. These contain respectively silver nitrate, mercuric nitrate, and mercurous nitrate. At the cathode of the cell $$P,0.216g$$ of $$Ag$$ was deposited. The weights of mercury deposited in the cathode of $$Q$$ and $$R$$ respectively are: (at wt. of $$Hg=200.59$$)

A
$$0.4012$$ and $$0.8024g$$

B
$$0.4012$$ and $$0.2006g$$

C
$$0.2006$$ and $$0.4012g$$

D
$$0.1003$$ and $$0.2006g$$

Solution

Cells are connected in series therefore same amount of current will Pass through them:
Faraday's second law:
$$\dfrac{W_1}{W_2}=\dfrac{E_1}{E_2}$$
(a) For Mercuric nitrate $$(Hg^{2+})$$
$$\dfrac{W_1}{(W_2)Ag}=\dfrac{E_1}{(E_2)Ag}$$
$$W_1=\dfrac{200.59}{2\times 108 }\times 0.216$$
$$=0.2006g$$

(b) For mercurous nitrate $$(Hg^+)$$
$$\dfrac{W_1}{(W_2)Ag}=\dfrac{E_1}{(E_2)Ag}$$ $$(\therefore E_{Ag}=108)$$
$$W_1=\dfrac{200.59}{108}\times 0.216$$
$$=0.4012g$$
Question 8
1 / -0

The time required for a current of $$3$$ amp. to decompose electrolytically $$18$$g of $$H_2O$$ is:

A
$$18$$ hour

B
$$36$$ hour

C
$$9$$ hour

D
$$18$$ seconds

Solution

According to Faraday first law $$W \propto Q$$
and $$W = \dfrac{Ewt\ } {F}$$
$$Ewt.=$$ Equivalent weight
$$F =$$ Faraday constant
$$W=$$ weight
$$Q =$$ $$current(i) \times time(t)$$

$$\dfrac {W} {Ewt} = \dfrac {it} {F}$$

$$\dfrac {18\times2}{18} =\dfrac {3\times t} {96500} (Ewt = \dfrac{M.w.}{V.F.})$$

$$\dfrac{2\times 96500}{3\times 3600} = t$$

$$ t =64333s \approx 18hr$$
The correct option is [A].
Question 9
1 / -0

Electrolysis can be used to determine atomic masses. A current of $$0.550\ A$$ deposits $$0.55$$. If a certain metal in $$100\ minutes$$. Calculate the atomic mass of the metal if eq. mass $$-$$ mole mass $$/ 3$$.

A
$$100$$

B
$$45.0$$

C
$$48.25$$

D
$$144.75$$

Solution

According to Faraday's law of Electrolysis
$$W=(\cfrac { M }{ nf } ).\cfrac { I.t }{ F } $$
$$0.55ℊ=(\cfrac { M }{ 3 } )\times 0.550\times (100\times 60)s$$
$$M=48.25ℊ$$
Thus atomic mass of metal is $$48.25$$
Question 10
1 / -0

Solution A,B and C of the same strong electrolyte offered resistances of $$50\ \Omega,\ 100\ \Omega,\ and\ 150\ \Omega$$ in a given conductivity cell. The resistance observed if they are mixed in a volume proportion which is reciprocal of their resistances and tested in the same conductivity cell would be:

A
$$67.3\ \Omega$$

B
$$81.8\ \Omega$$

C
$$100\ \Omega$$

D
$$300\ \Omega$$

Solution

Let value be $$v_1,\ v_2$$ and $$v_3$$
$$\dfrac {v_1}{R_1}+\dfrac {v_2}{R_2}+\dfrac {v_3}{R_3}=\dfrac {v_1 +v_2 +v_3}{Req}$$
$$R_1=52\ \Omega\ R_2=100\ \Omega \ R_3=150\ \Omega$$
$$v_1=\dfrac {1}{50}\ v_2=\dfrac {1}{100}\ v_3=\dfrac {1}{150}$$
$$\dfrac {v_1}{R_1}+\dfrac {v_2}{R_2}+\dfrac {v_3}{R_3}=\dfrac {v_1 +v_2 +v_3}{Req}$$
$$\dfrac {1}{50\times 50}+\dfrac {1}{100\times 100}+\dfrac {1}{150\times 150}=\dfrac {\dfrac {1}{50}+\dfrac {1}{100}+\dfrac {1}{150}}{Req}$$
$$\dfrac {36+9+4}{150\times 50\times 4}=\dfrac {\dfrac {6+3+2}{900}}{Req}$$
$$Req\ \dfrac {49}{150\times 150\times 4}=\dfrac {11}{300}$$
$$Req\ =\dfrac {3300}{49}=67.3\ \Omega$$

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Electrochemistry Test - 61

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Electrochemistry Test - 61

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Question 1

$$ 1.4 V $$

$$ 1.5 V $$

$$ 1.6 V $$

$$ 1.3 V $$

Solution

Question 2

+1280 cal

+640 cal

-1280 cal

-640 cal

Solution

Question 3

$$24,250$$

$$48,124$$

$$96,500$$

$$10,000$$

Question 4

$$+ 0.72$$

$$+ 1.54$$

$$+1.00$$

$$+ 1.35$$

Solution

Question 5

$$+0.44\ V$$

$$-0.44\ V$$

$$+0.48\ V$$

$$-0.48\ V$$

Solution

Question 6

-4.18 V and yes

+0.33 V and yes

+2.69 V and no

-2.69 V and no

Solution

Question 7

$$0.4012$$ and $$0.8024g$$

$$0.4012$$ and $$0.2006g$$

$$0.2006$$ and $$0.4012g$$

$$0.1003$$ and $$0.2006g$$

Solution

Question 8

$$18$$ hour

$$36$$ hour

$$9$$ hour

$$18$$ seconds

Solution

Question 9

$$100$$

$$45.0$$

$$48.25$$

$$144.75$$

Solution

Question 10

$$67.3\ \Omega$$

$$81.8\ \Omega$$

$$100\ \Omega$$

$$300\ \Omega$$

Solution

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