Electrochemistry Test

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Electrochemistry Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Cell reaction is spontaneous when

A
E^θRed is positive

B
ΔG^θ is positive

C
ΔG^θ is negative

D
E^θRed is negative.
Question 2
1 / -0

Three cell A, B and C has equilibrium constant in the ratio 1:4 : 9 respectively. Arrange the following cells in the order of increasing Gibbs free energy.

A
A>B>C

B
Cannot be answered

C
B>C>A

D
A
Solution
The correct answer is option A

The cells will be arranged as such 9 : 4: 1.
Explanation:
The equilibrium constant that is used here is represented with k.
The smaller the value of k the higher will be the value of energy according to Gibbs.
So, 1 has the highest Gibbs free energy while 9 has the least.
This is the energy of a chemical reaction.
It is represented by G.
Question 3
1 / -0

Gibbs free energy change for a cell reaction is positive what does it indicates?

A
cell will discharge easily

B
Cell reaction is spontaneous

C
Cell reaction is non spontaneous

D
Cell will work under standard conditions

Solution

The correct answer is Option C.
Reactions with a negative ∆G release energy, which means that they can proceed without an energy input (are spontaneous). In contrast, reactions with a positive ∆G need an input of energy in order to take place (are non-spontaneous). Reactions with a positive ∆H and negative ∆S are non-spontaneous at all temperatures.
Question 4
1 / -0

At equilibrium:

A
Cell potential’ E _cell‘ becomes zero

B
Equilibrium constant becomes equal to electrode potential

C
Equilibrium constant becomes zero

D
Cell potential ‘E _cell‘ becomes unity

Solution

The correct answer is option A
E cell is 0 in equilibrium, that is Ecathode becomes equal to E anode ……….
EZero cell becomes zero when both the electrodes are of the same metal but of different concentration i.e for concentration cell……
We know that EZero cell is - Ezerocathode -E zeroanode (since cathode and anode are same ) so EZero =0
Question 5
1 / -0

In the equation, ΔG° = – nF E° cell ; F is:

A
Boltzmann constant

B
Faraday’s constant

C
Gas constant

D
Universal gas constant

Solution

The correct answer is Option B.
The relationship between ΔG^o and E^o is given by the following equation: ΔG^o=−nFE^o. Here, n is the number of moles of electrons and F is the Faraday constant.
Question 6
1 / -0

Consider the cell reaction:
Cd(s) | Cd²⁺ (1.0 M) || Cu²⁺ (1.0 m) | Cu (s)
If we wish to make a cell with more positive voltage using the same substances, we should:

A
Increase [Cd²⁺] as well as [Cu²⁺] to 2.0 M

B
Increase only [Cu²⁺] to 2.0 M

C
Reduce only [Cd²⁺] to 0.1 M

D
Decreases [Cd²⁺] to 0.1M and increases [Cu²⁺] to 1.0M

Solution

The correct answer is Option D.
Redox reaction:
Cd(s)→Cd²⁺+2e
Cu²⁺+2e→Cu(s)
E_cell = E°_cell − (0.059/2) log ([Cd²⁺]/ [Cu²⁺])
Decreases [Cd²⁺] to 0.1M and increases [Cu²⁺] to 1.0M
Question 7
1 / -0

. The electrode potential at any concentration measured with respect to standard hydrogen electrode can be represented by:

A

B

C

D

Solution

The correct answer is Option B
Nernst equation is,

where, Q is the reaction quotient of the reaction
As, [M (s) ]=1
We get,
Question 8
1 / -0

. For an equation: Ni(s) + 2Ag⁺(aq) → Ni²⁺ (aq) + 2Ag(s) the Nernst equation is written as:

A

B

C

D

Solution

The correct answer is Option A.
Oxidation:
Ni_(s) ---> Ni²⁺_(aq) + 2e^-
Reduction:
2Ni⁺_(aq) + 2e^- ----> 2Aq_(s)
2e- are in the above reaction so;
n = 2
We know that
E_cell= E^o_cell – (RT/nF) lnK_c
= E^o_cell – (RT/nF) ln ([Ni]²⁺ / [Ag⁺]²
Question 9
1 / -0

Nernst equation for an electrode is based on the variation of electrode potential of an electrode with:

A
temperature only

B
Concentration of electrolyte only

C
Both a and b

D
Density of the electrodes

Solution

Nernst equation for an electrode is based on the variation of electrode potential of an electrode with temperature and concentration of electrolyte.
Question 10
1 / -0

The free energy change for the following cell reaction is given as :
2Au³⁺ (aq) + 3Cu (s) → 2Au (s) + 3Cu²⁺ (aq)

A
6 FE°cell

B
3 FE°cell

C
-2 FE°cell

D
-6 FE°cell

Solution

The correct answer is Option D.
E^O = E^O_Ca²⁺/ _Ca - E^O_Au²⁺/ _Au
= -2.87 - (1.50)
= -2.87 - 1.50
= -4.37 V
rG^O = nFE^O
= -6 FE^O