Question
Write the Nernst equation for the cell reaction in the Daniel cell. How will the ECell be affected when concentration of Zn2+ ions is increased?
✓
Answer
Daniell cell
$\text{Zn(s)}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu}|\text{Cu(s)}$
$\text{Anode}:\ \ \ \ \ \ \ \ \ \text{Zn(s)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{Zn}^{2+}+2\text{e}^{-}\\ \text{Cathode}:\ \ \ \ \ \text{Cu}^{2+}+2\text{e}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{Cu(s)}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ \text{Cell reaction}\ \ \text{Zn(s)}+\text{Cu}^{2+}\rightleftharpoons\text{Zn}^{2+}+\text{Cu(s)}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
$\text{Q}=\frac{\big[\text{Zn}^{2+}\big]}{\big[\text{Cu}^{2+}\big]}$
$\text{E}_{\text{Cell}}=\text{E}^\circ_{\text{cell}}-\frac{0.059}{2}\log\text{Q}=\frac{\big[\text{Zn}^{2+}\big]}{\big[\text{Cu}^{2+}\big]}$
Above equation shows that the cell potential will decrease with increase in the concentration of Zn2+ ion.
$\text{Zn(s)}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu}|\text{Cu(s)}$
$\text{Anode}:\ \ \ \ \ \ \ \ \ \text{Zn(s)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{Zn}^{2+}+2\text{e}^{-}\\ \text{Cathode}:\ \ \ \ \ \text{Cu}^{2+}+2\text{e}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{Cu(s)}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ \text{Cell reaction}\ \ \text{Zn(s)}+\text{Cu}^{2+}\rightleftharpoons\text{Zn}^{2+}+\text{Cu(s)}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
$\text{Q}=\frac{\big[\text{Zn}^{2+}\big]}{\big[\text{Cu}^{2+}\big]}$
$\text{E}_{\text{Cell}}=\text{E}^\circ_{\text{cell}}-\frac{0.059}{2}\log\text{Q}=\frac{\big[\text{Zn}^{2+}\big]}{\big[\text{Cu}^{2+}\big]}$
Above equation shows that the cell potential will decrease with increase in the concentration of Zn2+ ion.
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