(B) The Nernst equation for the given cell reaction is:$E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Here,$n = 2$,so the e

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(B) The Nernst equation for the given cell reaction is:$E_{cell} = E^o_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Here,$n = 2$,so the equation becomes:$E_{cell} = 1.10 - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$This is in the form of a linear equation $y = mx + c$,where $y = E_{cell}$,$x = log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$c = 1.10$,and the slope $m = -\frac{0.059}{2} = -0.0295$.Since the slope is negative,the graph of $E_{cell}$ versus $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$ will be a straight line with a negative slope and a $y$-intercept of $1.10 \, V$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Difficult

Which graph correctly correlates $E_{cell}$ as a function of concentrations for the cell (for different values of $M$ and $M'$):-
$Zn_{(s)} + Cu^{2+}(M) \to Zn^{2+}(M') + Cu_{(s)};$ $E^o_{cell} = 1.10 \, V$
$X$-axis : $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$Y$-axis : $E_{cell}$

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Class 12 Questions

Class 12

Chemistry Questions

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Electrochemistry

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Nernst equation and ECS

Find the $emf$ of the following cell reaction,given $E^0_{Cr^{3+}/Cr} = -0.74 \ V$ and $E^0_{Fe^{2+}/Fe} = -0.44 \ V$ at $25^{\circ} C$ for the cell: $Cr \ | \ Cr^{3+}(0.1 \ M) \ || \ Fe^{2+}(0.01 \ M) \ | \ Fe$. (in $V$)

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For the cell,$Mn_{(s)} | Mn^{2+}_{(aq)} (0.4 \, M) || Sn^{2+}_{(aq)} (0.04 \, M) | Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298 \, K$ in $kJ$. Given: $E^o_{Mn^{2+}/Mn} = -1.18 \, V$; $E^o_{Sn^{2+}/Sn} = -0.14 \, V$; $\frac{2.303RT}{F} = 0.06$.

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For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

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What must be the concentration of $Ag^{+}$ in an aqueous solution containing $Cu^{2+} = 1.0 \ M$ so that both the metals can be deposited on the cathode simultaneously? Given that $E^0_{Cu^{2+}/Cu} = 0.34 \ V$ and $E^0_{Ag^{+}/Ag} = 0.812 \ V$ at $T = 298 \ K$.

Medium

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For the redox reaction $Zn(s) + Cu^{2+}(0.1 \ M) \rightarrow Zn^{2+}(1 \ M) + Cu(s)$,given $E^o_{cell} = 1.10 \ V$,calculate the $E_{cell}$ value in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$) (in $V$)

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Which graph correctly correlates $E_{cell}$ as a function of concentrations for the cell (for different values of $M$ and $M'$):-$Zn_{(s)} + Cu^{2+}(M) \to Zn^{2+}(M') + Cu_{(s)};$ $E^o_{cell} = 1.10 \, V$$X$-axis : $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$Y$-axis : $E_{cell}$

Similar Questions

Find the $emf$ of the following cell reaction,given $E^0_{Cr^{3+}/Cr} = -0.74 \ V$ and $E^0_{Fe^{2+}/Fe} = -0.44 \ V$ at $25^{\circ} C$ for the cell: $Cr \ | \ Cr^{3+}(0.1 \ M) \ || \ Fe^{2+}(0.01 \ M) \ | \ Fe$. (in $V$)

For the cell,$Mn_{(s)} | Mn^{2+}_{(aq)} (0.4 \, M) || Sn^{2+}_{(aq)} (0.04 \, M) | Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298 \, K$ in $kJ$. Given: $E^o_{Mn^{2+}/Mn} = -1.18 \, V$; $E^o_{Sn^{2+}/Sn} = -0.14 \, V$; $\frac{2.303RT}{F} = 0.06$.

For the galvanic cell,$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

What must be the concentration of $Ag^{+}$ in an aqueous solution containing $Cu^{2+} = 1.0 \ M$ so that both the metals can be deposited on the cathode simultaneously? Given that $E^0_{Cu^{2+}/Cu} = 0.34 \ V$ and $E^0_{Ag^{+}/Ag} = 0.812 \ V$ at $T = 298 \ K$.

For the redox reaction $Zn(s) + Cu^{2+}(0.1 \ M) \rightarrow Zn^{2+}(1 \ M) + Cu(s)$,given $E^o_{cell} = 1.10 \ V$,calculate the $E_{cell}$ value in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$) (in $V$)

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Which graph correctly correlates $E_{cell}$ as a function of concentrations for the cell (for different values of $M$ and $M'$):-
$Zn_{(s)} + Cu^{2+}(M) \to Zn^{2+}(M') + Cu_{(s)};$ $E^o_{cell} = 1.10 \, V$
$X$-axis : $log_{10} \frac{[Zn^{2+}]}{[Cu^{2+}]}$,$Y$-axis : $E_{cell}$

For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$