Consider the following ${E^0}$ values:
${E^0}_{Fe^{3+}/Fe^{2+}} = + 0.77 \ V$
${E^0}_{Sn^{2+}/Sn} = - 0.14 \ V$
Under standard conditions,the potential for the reaction $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ is ............ $V$.
${E^0}_{Fe^{3+}/Fe^{2+}} = + 0.77 \ V$
${E^0}_{Sn^{2+}/Sn} = - 0.14 \ V$
Under standard conditions,the potential for the reaction $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ is ............ $V$.
- A$0.91$
- B$1.40$
- C$1.68$
- D$0.63$
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Similar Questions
Given the standard electrode potentials $E^{\circ}_{Fe^{3+}/Fe^{2+}} = +0.77 \ V$ and $E^{\circ}_{Sn^{2+}/Sn} = -0.14 \ V$,calculate the standard cell potential $E^{\circ}_{cell}$ for the reaction: $Sn_{(s)} + 2Fe^{3+}_{(aq)} \rightarrow 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ (in $V$)
Easy
View SolutionMatch the following:
List-$I$ List-$II$ $(A)$ Potential of hydrogen electrode at $pH = 10$ $(I)$ $0.76 \ V$ $(B)$ $Cu^{2+} | Cu$ $(II)$ $0.059$ $(C)$ $Zn | Zn^{2+}$ $(III)$ $-0.591 \ V$ $(D)$ $\frac{2.303 RT}{F}$ $(IV)$ $0.337 \ V$ $(V)$ $-0.76 \ V$
$A$ $B$ $C$ $D$
$(a)$ $(III)$ $(I)$ $(II)$ $(V)$
$(b)$ $(II)$ $(V)$ $(I)$ $(IV)$
$(c)$ $(III)$ $(IV)$ $(I)$ $(II)$
$(d)$ $(V)$ $(I)$ $(IV)$ $(II)$
| List-$I$ | List-$II$ |
|---|---|
| $(A)$ Potential of hydrogen electrode at $pH = 10$ | $(I)$ $0.76 \ V$ |
| $(B)$ $Cu^{2+} | Cu$ | $(II)$ $0.059$ |
| $(C)$ $Zn | Zn^{2+}$ | $(III)$ $-0.591 \ V$ |
| $(D)$ $\frac{2.303 RT}{F}$ | $(IV)$ $0.337 \ V$ |
| $(V)$ $-0.76 \ V$ |
$A$ $B$ $C$ $D$
$(a)$ $(III)$ $(I)$ $(II)$ $(V)$
$(b)$ $(II)$ $(V)$ $(I)$ $(IV)$
$(c)$ $(III)$ $(IV)$ $(I)$ $(II)$
$(d)$ $(V)$ $(I)$ $(IV)$ $(II)$
Given that $E^o_{K^{+}/K} = -2.93 \ V$,$E^o_{Fe^{2+}/Fe} = -0.44 \ V$,$E^o_{Zn^{2+}/Zn} = -0.76 \ V$,and $E^o_{Cu^{2+}/Cu} = 0.34 \ V$. Based on this data,which of the following is the strongest reducing agent?
DifficultAIIMS 2011
View SolutionThe $E^o$ for half cells $Fe/Fe^{2+}$ and $Cu/Cu^{2+}$ are $-0.44 \, V$ and $+0.32 \, V$ respectively. Then
Easy
View SolutionThe $E^{0}_{Red}$ values of $A, B,$ and $C$ are $+0.5 \ V, -3.0 \ V,$ and $-1.2 \ V$ respectively. The order of their reducing power is:
Easy
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