The cell potential for the given cell at 298 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The cell reaction is: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2 H^{+}_{(aq)} + Cu_{(s)}$Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} -

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) The cell reaction is: $H_{2(g)} + Cu^{2+}_{(aq)} \rightarrow 2 H^{+}_{(aq)} + Cu_{(s)}$Using the Nernst equation: $E_{cell} = E^{\ominus}_{cell} - \frac{0.06}{n} \log \frac{[H^{+}]^2}{[Cu^{2+}]}$Here,$n = 2$,$E_{cell} = 0.31 \, V$,$E^{\ominus}_{cell} = E^{\ominus}_{Cu^{2+}/Cu} - E^{\ominus}_{H^{+}/H_2} = 0.34 - 0 = 0.34 \, V$.Given $pH = 3$,so $[H^{+}] = 10^{-3} \, M$.Substituting the values: $0.31 = 0.34 - \frac{0.06}{2} \log \frac{(10^{-3})^2}{[Cu^{2+}]}$$-0.03 = -0.03 \log \frac{10^{-6}}{[Cu^{2+}]}$$1 = \log \frac{10^{-6}}{[Cu^{2+}]}$$10^1 = \frac{10^{-6}}{[Cu^{2+}]}$$[Cu^{2+}] = 10^{-7} \, M$.Comparing with $10^{-x} \, M$,we get $x = 7$.

Similar Questions

Calculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)

The standard cell potential of a cell reaction involving the transfer of $2$ electrons at $25^{\circ}C$ is $0.295 \ V$. What is its equilibrium constant at $25^{\circ}C$?

For the cell $Co|Co^{2+}(C_2)||Co^{2+}(C_1)|Co$,$\Delta G$ is negative if

Calculate the equilibrium constant for the cell reaction at $298 \ K$: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,given $E^{0}_{cell} = 0.46 \ V$.

For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module