Question

Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: 2H+(aq) + Ni(s) H2(g) + Ni2+(aq) Answer: kJ K for this reaction would be ____(less/greater) than one.

Answer 1

The half reactions are shown below.

1. 2H+(aq) + 2e- ……………..> H2(g)             Er = 0.0 V
2. Ni2+(aq) + 2e- ……………> Ni(s)                Eo = -0.236 V

Now (1) – (2) =>

2H+(aq) + Ni(s) …………..> H2(g) + Ni2(aq)            E = 0.236 V

Now, dG = -nFE ,

Here n = 2 mol, F = 96500 J/V.mol, E = 0.236 V

dG = - (2 x 96500 x 0.236) J = - 45548 J = - 45.548 kJ

We have E = (0.0592/n) logK

Where K is the equilibrium constant.

Using the above relation,

logK = (0.236 x 2)/0.0592 = 7.973 => K = 9.397 x 10⁷

Hence, the K value for this reaction is greater than one.