If Zn/Zn^{+2} electrode is diluted 100 times, | vedclass

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

Question

(B) The electrode potential for $Zn^{+2}/Zn$ is given by the Nernst equation: $E = E^o - \frac{0.059}{n} \log \frac{1}{[Zn^{+2}]}$.For $Zn^{+2} + 2e^-

Vedclass · Accepted Answer

decrease of $59 \ mV$

Vedclass · Answer

(B) The electrode potential for $Zn^{+2}/Zn$ is given by the Nernst equation: $E = E^o - \frac{0.059}{n} \log \frac{1}{[Zn^{+2}]}$.For $Zn^{+2} + 2e^- ightarrow Zn$,$n = 2$.$E = E^o - \frac{0.059}{2} \log \frac{1}{[Zn^{+2}]} = E^o + \frac{0.059}{2} \log [Zn^{+2}]$.When the solution is diluted $100$ times,the concentration $[Zn^{+2}]$ becomes $[Zn^{+2}] / 100$.The new potential $E' = E^o + \frac{0.059}{2} \log \frac{[Zn^{+2}]}{100} = E^o + \frac{0.059}{2} (\log [Zn^{+2}] - \log 100)$.$E' = E^o + \frac{0.059}{2} \log [Zn^{+2}] - \frac{0.059}{2} 	imes 2$.$E' = E - 0.059 \ V = E - 59 \ mV$.Thus,the electromotive force decreases by $59 \ mV$.

If $Zn/Zn^{+2}$ electrode is diluted $100$ times,then the change in electromotive force will be:

Similar Questions

The reduction potential of a half cell consisting of a $Pt$ electrode immersed in $2.0 \ M \ Fe^{2+}$ and $0.02 \ M \ Fe^{3+}$ solution (in $V$) is. Given: $\left(\frac{2.303 \ RT}{F} = 0.059, E^0_{Fe^{3+} \mid Fe^{2+}} = 0.771 \ V\right)$

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

Calculate the cell potential for $Cr_{(s)} | Cr^{3+} (0.1 \, M) || Fe^{2+} (0.01 \, M) | Fe_{(s)}$ at $298 \, K$. Given $E^{\circ}_{Cr^{3+}/Cr} = -0.74 \, V$ and $E^{\circ}_{Fe^{2+}/Fe} = -0.44 \, V$. (in $, V$)

For the cell reaction $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$,what happens when $H_2SO_4$ is added to the cathode compartment?

$Zn_{(s)} + Cl_2(1 \ atm) \to Zn^{2+} + 2Cl^-$. The $E^0_{cell}$ of the cell is $2.12 \ V$. To increase $E_{cell}$:

Vedclass Test Series

Exam Paper Generator

Online Exam Module