By how much would the oxidising power of the $(MnO_4^- / Mn^{2+})$ couple change if the $H^{+}$ ion concentration is increased up to $100$ times at $25 ^oC$?

  • A
    Increase by $189 \ mV$
  • B
    Decrease by $189 \ mV$
  • C
    Will increase by $19 \ mV$
  • D
    Will decrease by $19 \ mV$

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The magnitude of the change in oxidising power of the $MnO_4^- / Mn^{2+}$ couple is $x \times 10^{-4} \, V$,if the $H^{+}$ concentration is decreased from $1 \, M$ to $10^{-4} \, M$ at $25^{\circ} C$. (Assume concentration of $MnO_4^-$ and $Mn^{2+}$ to be same on change in $H^{+}$ concentration). The value of $x$ is ....... .
(Rounded off to the nearest integer)
$[\text{Given} : \frac{2.303 RT}{F} = 0.059]$

For a redox reaction,$Oxi + ne^- \rightarrow Red$,what is the correct form of the Nernst equation?

$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be

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At $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)

At $298 \, K$,find out the $emf$ for the cell: $Al_{(s)} | Al^{+3} (0.1 \, M) || Fe^{+2} (0.001 \, M) | Fe_{(s)}$. Given $E^o_{Al^{+3}/Al} = -1.66 \, V$ and $E^o_{Fe^{+2}/Fe} = -0.44 \, V$. (in $, V$)

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