The standard potential of a Co^{2+}|Co electr | vedclass

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Question

(B) The cell reaction is $Ti^{2+} + Co^{2+} \rightarrow Ti^{3+} + Co$. The standard cell potential is given by $E_{cell}^{o} = E_{cathode}^{o} - E_{an

Vedclass · Accepted Answer

$0.37$

Vedclass · Answer

(B) The cell reaction is $Ti^{2+} + Co^{2+} \rightarrow Ti^{3+} + Co$. The standard cell potential is given by $E_{cell}^{o} = E_{cathode}^{o} - E_{anode}^{o}$. Here,the cathode is the $Co^{2+}|Co$ electrode and the anode is the $Ti^{2+}|Ti^{3+}$ electrode. $E_{cell}^{o} = E_{Co^{2+}|Co}^{o} - E_{Ti^{3+}|Ti^{2+}}^{o}$. Substituting the given values: $0.09 \ V = -0.28 \ V - E_{Ti^{3+}|Ti^{2+}}^{o}$. $E_{Ti^{3+}|Ti^{2+}}^{o} = -0.28 \ V - 0.09 \ V = -0.37 \ V$. The standard oxidation potential is the negative of the standard reduction potential: $E_{ox}^{o} = -E_{red}^{o} = -(-0.37 \ V) = 0.37 \ V$.

$Zn \rightarrow Zn^{2+} + 2e^-$	$E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$	$E^{\circ} = +0.41 \ V$

The standard potential of a $Co^{2+}|Co$ electrode is $-0.28 \ V$ and the standard potential of the cell $Pt|Ti^{2+}_{(aq)}, Ti^{3+}_{(aq)}||Co^{2+}_{(aq)}|Co_{(s)}$ is $0.09 \ V$. What is the standard oxidation potential of the $Ti^{2+}_{(aq)}|Ti^{3+}_{(aq)}$ electrode? ............. $V$

Similar Questions

The standard electrode potential for the half-cell reactions are
$Zn^{2+} + 2e^{-} \longrightarrow Zn ; E^{\circ} = -0.76 \ V$
$Fe^{2+} + 2e^{-} \longrightarrow Fe ; E^{\circ} = -0.44 \ V$
The $emf$ of the cell reaction,
$Fe^{2+} + Zn \longrightarrow Zn^{2+} + Fe$ is

The oxidation potentials of $Zn, Cu, Ag, H_2$ and $Ni$ are $0.76, -0.34, -0.80, 0$ and $0.25 \ V$ respectively. Which reaction provides the maximum voltage?

The standard reduction potential $E^{\circ}$ for half reactions are
$Zn \rightarrow Zn^{2+} + 2e^-$ $E^{\circ} = +0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^-$ $E^{\circ} = +0.41 \ V$

The $EMF$ of the cell reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is

The standard $E^{\circ}_{red}$ values of $A, B$ and $C$ are $+0.68 \ V, -2.54 \ V$ and $-0.50 \ V$ respectively. The order of their reducing power is:

Given $E^{0}_{Fe^{2+} | Fe} = -0.44 \, V$ and $E^{0}_{Sn^{2+} | Sn} = -0.14 \, V$,calculate $E^{0}_{cell}$ for the reaction $Fe^{2+} + Sn \rightarrow Fe + Sn^{2+}$.

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