The emf of the cell Tl_{(s)} | Tl^{+}_{(aq)} | vedclass

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Question

(D) The cell reaction is: $2Tl_{(s)} + Cu^{2+}_{(aq)} \rightarrow 2Tl^{+}_{(aq)} + Cu_{(s)}$.According to the Nernst equation: $E_{cell} = E^{0}_{cell

Vedclass · Accepted Answer

Increase the concentration of $Cu^{2+}$ ions and decrease the concentration of $Tl^{+}$ ions.

Vedclass · Answer

(D) The cell reaction is: $2Tl_{(s)} + Cu^{2+}_{(aq)} \rightarrow 2Tl^{+}_{(aq)} + Cu_{(s)}$.According to the Nernst equation: $E_{cell} = E^{0}_{cell} - \frac{0.0591}{n} \log \frac{[Tl^{+}]^2}{[Cu^{2+}]}$.To increase $E_{cell}$,the value of the logarithmic term $\frac{[Tl^{+}]^2}{[Cu^{2+}]}$ must decrease.This can be achieved by increasing the concentration of the reactant $[Cu^{2+}]$ or decreasing the concentration of the product $[Tl^{+}]$.Therefore,both increasing $[Cu^{2+}]$ and decreasing $[Tl^{+}]$ will increase the $emf$ of the cell.

The $emf$ of the cell $Tl_{(s)} | Tl^{+}_{(aq)} (0.0001 \ M) || Cu^{2+}_{(aq)} (0.01 \ M) | Cu_{(s)}$ is $0.83 \ V$. How can the $emf$ of this cell be increased?

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