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?

Similar Questions

The cell potential for the following cell notation is approximately
$M_{(s)} | M^{3+}(aq, 0.01 \ M) || N^{2+}(aq, 0.1 \ M) | N_{(s)}$
$E_{M^{3+} / M}^0 = 0.6 \ V$ and $E_{N^{2+} / N}^0 = 0.1 \ V$ (in $V$)

The e.m.f. of the cell in which the following reaction $Zn_{(s)} + Ni^{2+}(a = 1.0) \rightleftharpoons Zn^{2+}(a = 10) + Ni_{(s)}$ occurs,is found to be $0.5105 \ V$ at $298 \ K$. The standard e.m.f. of the cell is ............ $V$.

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For the redox reaction $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$ taking place in a cell,$E_{cell}^o$ is $1.10 \ V$. $E_{cell}$ for the cell will be ............ $V$ $\left( 2.303 \frac{RT}{F} = 0.0591 \right)$

The standard emf for the cell $Cd_{(s)}|Cd^{2+}_{(aq)}(1 \ M)||Cu^{2+}_{(aq)}(1 \ M)|Cu_{(s)}$ is $0.74 \ V$. If the concentration of $Cd^{2+}_{(aq)}$ and $Cu^{2+}_{(aq)}$ both decrease by $10$ times at $298 \ K$,calculate the emf of the cell.

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