Question

The solution of CuSO${}_4$ in which copper rod is immersed is diluted to 10 times, the reduction electrode potential:

• A. increase by $0.030V$
• B. decreases by $0.030V$
• C. increases by $0.059V$
• D. decreases by $0.059V$

Answer 1

Answer: (B)

decreases by $0.030V$

According to Nernst equations, $E=E^o-\left(0.0591/n\right)logQ$

$E_{R{P}_1}=E^o+\frac{0.059}{2}log[C{u}^{2+}{]}_1$.........(1);

$[C{u}^{2+}{]}_2=C_1\times \frac{1}{10}$

So now, $E_{R{P}_2}=E^o+\frac{0.059}{2}log{C}_1\times \frac{1}{10}$ .........(2)

So, $E_2-E_1=-0.030V$

So reduction electrode potential decreases by 0.030 V