An electrochemical cell consists of the following two redox couples, $M^{x+}(aq)/M(s)$ $[E^{\ominus}_{red} = +0.15 \text{ V}]$ and $Fe^{3+}(aq)/Fe(s)$ $[E^{\ominus}_{red} = -0.036 \text{ V}]$. The cell $EMF$ is recorded to be $0.2057 \text{ V}$. If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of $x$ is . . . . . . . (Nearest integer) [Given: $M$ is a $p$-block metal and $\frac{2.303RT}{F} = 0.059 \text{ V}$]
- A$1$
- B$2$
- C$3$
- D$4$
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In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:
Medium
View SolutionConsider the following cell reaction:
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$
At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:
$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$
$2 Fe^{3+}_{(aq)} + 2 I^{-}_{(aq)} \rightleftharpoons 2 Fe^{2+}_{(aq)} + I_{2(s)}$
At $298 \ K$,the cell emf is $0.237 \ V$. The equilibrium constant for the reaction is $10^x$. The value of $x$ is:
$(F = 96500 \ C \ mol^{-1}; R = 8.3 \ J \ K^{-1} \ mol^{-1})$
In 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,$RT/F = 0.059$)
Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)
Easy
View SolutionThe $EMF$ of the cell $Cr|Cr^{3+}(0.1 \ M) \| Fe^{2+}(0.01 \ M)|Fe$ is: $(E^{\circ}_{Cr^{3+}|Cr} = -0.75 \ V, E^{\circ}_{Fe^{2+}|Fe} = -0.45 \ V)$ (in $V$)
Medium
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