(A) The cell reaction is: $Mg_{(s)} + 2H^+_{(aq)} \rightarrow Mg^{2+}_{(aq)} + H_{2(g)}$.Given $E^\circ_{cell} = 2.36 \ V$ at standard conditions ($1

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Anode: $Mg \rightarrow Mg^{2+} + 2e^-$,Cathode: $2H^+ + 2e^- \rightarrow H_2$,$E^\circ_{Mg^{2+}/Mg} = -2.36 \ V$

(A) The cell reaction is: $Mg_{(s)} + 2H^+_{(aq)} \rightarrow Mg^{2+}_{(aq)} + H_{2(g)}$.Given $E^\circ_{cell} = 2.36 \ V$ at standard conditions ($1 \ M$ concentration and $1 \ bar$ pressure).The cell potential is given by $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$.Here,the cathode is the Standard Hydrogen Electrode $(SHE)$,so $E^\circ_{H^+/H_2} = 0.00 \ V$.Therefore,$2.36 \ V = 0.00 \ V - E^\circ_{Mg^{2+}/Mg}$.$E^\circ_{Mg^{2+}/Mg} = -2.36 \ V$.

Class 12 Chemistry Electrochemistry Electrode potential and ECell

Medium

If the cell potential of the cell at $298 \ K$ is $2.36 \ V$,write the cell reaction and calculate the standard electrode potential of the $Mg^{2+} \mid Mg$ half-cell.
$Mg_{(s)} \mid Mg^{2+}_{(1 \ M)} \parallel H^{+}_{(1 \ M)} \mid H_{2(g)} (1 \ bar) \mid Pt_{(s)}$

A
Anode: $Mg \rightarrow Mg^{2+} + 2e^-$,Cathode: $2H^+ + 2e^- \rightarrow H_2$,$E^\circ_{Mg^{2+}/Mg} = -2.36 \ V$
B
Anode: $Mg \rightarrow Mg^{2+} + 2e^-$,Cathode: $2H^+ + 2e^- \rightarrow H_2$,$E^\circ_{Mg^{2+}/Mg} = +2.36 \ V$
C
Anode: $Mg^{2+} + 2e^- \rightarrow Mg$,Cathode: $H_2 \rightarrow 2H^+ + 2e^-$,$E^\circ_{Mg^{2+}/Mg} = -2.36 \ V$
D
Anode: $Mg \rightarrow Mg^{2+} + 2e^-$,Cathode: $2H^+ + 2e^- \rightarrow H_2$,$E^\circ_{Mg^{2+}/Mg} = -1.18 \ V$

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Electrode potential and ECell

Calculate $\Delta G^{\circ}$ for the cell: $Sn_{(s)} | Sn^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ at $25^{\circ} C$ given that $E^{\circ}_{cell} = 0.90 \ V$. (in $kJ$)

MediumMHT CET 2023

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Given,for $Sn^{4+} / Sn^{2+}$,standard reduction potential is $0.15 \ V$ and for $Au^{3+} / Au$,standard reduction potential is $1.5 \ V$. For the reaction,$3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$,the value of $E_{\text{cell}}^{\circ}$ is:

MediumMHT CET 2010

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Medium

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If in a galvanic cell,the hydrogen half-cell is the cathode on the right side,what is the sign of the standard electrode potential of such a galvanic cell? Give an example.

Easy

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For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E^\circ_{cell}$ is $1.89 \ V$. If $E^\circ_{Co^{2+}/Co} = -0.28 \ V$,then $E^\circ_{Ce^{4+}/Ce^{3+}}$ is equal to:

Medium

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If the cell potential of the cell at $298 \ K$ is $2.36 \ V$,write the cell reaction and calculate the standard electrode potential of the $Mg^{2+} \mid Mg$ half-cell.$Mg_{(s)} \mid Mg^{2+}_{(1 \ M)} \parallel H^{+}_{(1 \ M)} \mid H_{2(g)} (1 \ bar) \mid Pt_{(s)}$

Similar Questions

Calculate $\Delta G^{\circ}$ for the cell: $Sn_{(s)} | Sn^{2+}_{(1M)} || Ag^{+}_{(1M)} | Ag_{(s)}$ at $25^{\circ} C$ given that $E^{\circ}_{cell} = 0.90 \ V$. (in $kJ$)

Given,for $Sn^{4+} / Sn^{2+}$,standard reduction potential is $0.15 \ V$ and for $Au^{3+} / Au$,standard reduction potential is $1.5 \ V$. For the reaction,$3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$,the value of $E_{\text{cell}}^{\circ}$ is:

For the half-cell $Zn^{2+} | Zn$,the standard electrode potential $E^{\circ}$ is $-0.76 \ V$. What is the $e.m.f.$ of the cell $Zn_{(s)} | Zn^{2+}_{(aq)} (1 \ M) || 2H^{+}_{(aq)} (1 \ M) | H_{2(g)} (1 \ atm)$ in $V$?

If in a galvanic cell,the hydrogen half-cell is the cathode on the right side,what is the sign of the standard electrode potential of such a galvanic cell? Give an example.

For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E^\circ_{cell}$ is $1.89 \ V$. If $E^\circ_{Co^{2+}/Co} = -0.28 \ V$,then $E^\circ_{Ce^{4+}/Ce^{3+}}$ is equal to:

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If the cell potential of the cell at $298 \ K$ is $2.36 \ V$,write the cell reaction and calculate the standard electrode potential of the $Mg^{2+} \mid Mg$ half-cell.
$Mg_{(s)} \mid Mg^{2+}_{(1 \ M)} \parallel H^{+}_{(1 \ M)} \mid H_{2(g)} (1 \ bar) \mid Pt_{(s)}$