Question

If the solution of the ${CuSO}_4$ in which copper rod is immersed is diluted to $10$ times, the electrode reduction potential. $(T=298K)$ :

• A. increases by $0.030volt$
• B. decreases by $0.030volt$
• C. increases by $0.059volt$
• D. decreases by $0.0059volt$

Answer 1

The correct option is B decreases by $0.030volt$

${Cu}^{2+}+{2e}^{-}⟶Cu$
$V_1=1,C_1=1M$
$V_2=10,C_2=0.1M$
$E_{cell}=E^o-\frac{0.0591}{2}\log \frac{1}{1}{E}_{cell}=E^o$
$E_{cell}=E^o-\frac{0.0591}{2}\log \frac{1}{0.1}$
$=E^o-0.059/2=E^o-0.030$