Consider the following electrochemical cell at $298 \ K$:
$Pt | HSnO_2^-(aq) | Sn(OH)_6^{2-}(aq) || Bi_2O_3(s) | Bi(s)$.
If the reaction quotient at a given time is $10^6$, then the cell $EMF$ $(E_{\text{cell}})$ is . . . . . . $\times 10^{-1} \ V$ (Nearest integer).
Given the standard half-cell reduction potential as
$E^0_{Bi_2O_3/Bi, OH^-} = -0.44 \ V$ and
$E^0_{Sn(OH)_6^{2-}/HSnO_2^-, OH^-} = -0.90 \ V$.
$Pt | HSnO_2^-(aq) | Sn(OH)_6^{2-}(aq) || Bi_2O_3(s) | Bi(s)$.
If the reaction quotient at a given time is $10^6$, then the cell $EMF$ $(E_{\text{cell}})$ is . . . . . . $\times 10^{-1} \ V$ (Nearest integer).
Given the standard half-cell reduction potential as
$E^0_{Bi_2O_3/Bi, OH^-} = -0.44 \ V$ and
$E^0_{Sn(OH)_6^{2-}/HSnO_2^-, OH^-} = -0.90 \ V$.
- A$2$
- B$3$
- C$4$
- D$5$
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Calculate the $E^o_{cell}$ for the following reaction: $Cu^{+2}_{(aq)} + Sn^{+2}_{(aq)} \rightarrow Cu_{(s)} + Sn^{+4}_{(aq)}$,given that the equilibrium constant $K_c = 10^6$. (in $V$)
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View SolutionFor the cell, $Pt | Cl_{2(g)} (0.4 \ bar) | Cl^{-} (aq.) (0.1 \ M) || Cl^{-} (aq.) (0.01 \ M) | Cl_{2(g)} (0.2 \ bar) | Pt$, the measured potential at $298 \ K$ is .............. $V$.
Medium
View SolutionIn the electrochemical cell $:$
$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\,M)\,|\,Cu$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\,M$ and that of $CuSO_4$ changed to $0.01\,M,$ the $emf$ changes to $E_2.$ From the followings,which one is the relationship between $E_1$ and $E_2$ $?$ (Given,$RT/F = 0.059$)
$Zn \,|\,ZnSO_4\,(0.01\,M)\,||\,CuSO_4\,(1.0\,M)\,|\,Cu$
the $emf$ of this Daniell cell is $E_1.$ When the concentration of $ZnSO_4$ is changed to $1.0\,M$ and that of $CuSO_4$ changed to $0.01\,M,$ the $emf$ changes to $E_2.$ From the followings,which one is the relationship between $E_1$ and $E_2$ $?$ (Given,$RT/F = 0.059$)
Assume a cell with the following reaction:
$Cu_{(s)} + 2 Ag^{+} (1 \times 10^{-3} \, M) \rightarrow Cu^{2+} (0.250 \, M) + 2 Ag_{(s)}$
$E_{Cell}^{\ominus} = 2.97 \, V$
$E_{cell}$ for the above reaction is $.... \, V.$ (Nearest integer)
[Given: $\log 2.5 = 0.3979, T = 298 \, K]$
$Cu_{(s)} + 2 Ag^{+} (1 \times 10^{-3} \, M) \rightarrow Cu^{2+} (0.250 \, M) + 2 Ag_{(s)}$
$E_{Cell}^{\ominus} = 2.97 \, V$
$E_{cell}$ for the above reaction is $.... \, V.$ (Nearest integer)
[Given: $\log 2.5 = 0.3979, T = 298 \, K]$
DifficultJEE MAIN 2021
View SolutionCalculate $E_{cell}^{\circ}$ if the equilibrium constant for the following reaction is $1.2 \times 10^6$.
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)
$2 Cu_{(aq)}^{+} \longrightarrow Cu_{(aq)}^{2+} + Cu_{(s)}$ (in $V$)
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