Calculate for an electrochemical cell reaction that occurs under basic aqueous conditions based on the following two half-reactions for which the standard reduction potentials are given.Use the smallest whole-number coefficients possible when balancing the overall reaction. Cd(OH)₂ + 2 e^-$\to$ Cd + 2 OH^- $\quad $ $\quad $ $\quad $-0.824 V NiO(OH)+ H₂O + e^-$\to$ Ni(OH)₂ + OH^- $\quad $+1.32 V

Question

Quizplus · Accepted Answer

The given half-reactions are:
Cd(OH)2 + 2e- → Cd + 2OH-     E° = -0.824 V
Ni(OH)2 + e- → Ni(OH) + OH-  E° = +0.320 V

To balance the electrons, we need to multiply the first half-reaction by 2 and combine with the second half-reaction as follows:

2Cd(OH)2 + Ni(OH)2 + 2e- → 2Cd + Ni(OH) + 4OH-
The overall E° can be calculated as:
E°cell = E°cathode - E°anode
= 0.320 - (-0.824)
= +1.144 V

The n value in the Nernst equation is the number of electrons transferred in the balanced equation, which is 2 in this case.

Using the Nernst equation: Ecell = E°cell - (0.0592/2) log(Q)
where Q = [Ni(OH)][Cd]^2/[Cd(OH)2]^2[OH]^4

At basic pH, the concentration of OH- is higher than H+, so [H+] can be assumed to be 1.0 x 10^-14 and [OH-] can be calculated as 1.0 x 10^-14/[OH-].

Plugging in the values, we get:
Ecell = 1.144 - (0.0592/2) log[(1.0 x 10^-14/[OH-]) Ni(OH)(Cd)^2 / (Cd(OH)2)^2(OH)^4]
= 1.144 - (0.0296) log[(Ni(OH)(Cd)^2] / [(Cd(OH)2)^2(OH)^4]

Simplifying further, we get:
Ecell = 1.144 + (0.0296) log[(OH)^4(Ni)(Cd)^2] - 2(0.0296) log[Cd(OH)2]

Substituting the given values, we get:
Ecell = 1.144 + (0.0296) log[(1.0 x 10^-14)^4(1)(10^-5)^2] - 2(0.0296) log[(0.1)^2]
= -0.538 V

The negative value of Ecell indicates that the reaction is not spontaneous under standard conditions. The value of ΔG° can be calculated using the equation ΔG° = -nFE°cell:

ΔG° = -2 x (-0.538) x F
= +209.67 kJ/mol or +414 kJ (rounded to 3 significant figures)

Therefore, the correct answer is D: F1F1F1S1F1F1F10414 kJ.

Calculate for an Electrochemical Cell Reaction That Occurs Under →\to→

Calculate for an Electrochemical Cell Reaction That Occurs Under $\to$