The emf of cell Tl | Tl^+_{(0.001M)} || Cu^{2 | 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^o_{cell} - \f

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increasing concentration of $Cu^{2+}$ ions

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(D) The cell reaction is: $2Tl_{(s)} + Cu^{2+}_{(aq)} \rightarrow 2Tl^+_{(aq)} + Cu_{(s)}$According to the Nernst equation:$E_{cell} = E^o_{cell} - \frac{0.0591}{2} \log \frac{[Tl^+]^2}{[Cu^{2+}]}$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,increasing the concentration of $Cu^{2+}$ ions will increase the $E_{cell}$.

The $emf$ of cell $Tl | Tl^+_{(0.001M)} || Cu^{2+}_{(0.01M)} | Cu$ is $0.83 \ V$ at $298 \ K$. It could be increased by:

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