Question

At what concentration of $C{u}^{2+}$ in a solution of $CuS{O}_4$ will the electrode potential be zero at ${25}^{\circ }\text{C}$? Given: $E_{cu|c{u}^{2+}}^o=-0.34\text{V}$

• A. $3.01\times {10}^{-12}$
• B. $2.98\times {10}^{-15}$
• C. $3.98\times {10}^{-20}$
• D. $2.89\times {10}^{-18}$

Answer 1

The correct option is A $3.01\times {10}^{-12}$

$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right);E_{cell}^o=0.34\text{V}$

$E_{cell}=E_{cell}^o-\frac{0.059}{n}\log \left(\frac{1}{\left[c{u}^{+2}\right]}\right)$
According to question
$0=0.34-\frac{0.059}{2}\log \left(\frac{1}{\left[C{u}^{2+}\right]}\right)$

$\Rightarrow \frac{0.34}{0.0295}=\log \left(\frac{1}{\left[c{u}^{+2}\right]}\right)$

$\Rightarrow +11.52=\log \left(\frac{1}{\left[c{u}^{+2}\right]}\right)$

$\Rightarrow \left[C{u}^{+2}\right]={10}^{-11.52}$

$\Rightarrow \left[C{u}^{+2}\right]=3.01\times {10}^{-12}\text{M}$