Every question from past board papers on EMF, Nernst equation and applications, newest paper first. Open any question for its full worked answer.
Given below are half cell reactions :
\(MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O\),
\(E^\circ_{Mn^{2+}/MnO_4^-} = -1.510\) V
\(\frac{1}{2}O_2 + 2H^+ + 2e^- \to H_2O\),
\(E^\circ_{O_2/H_2O} = +1.223\) V
Will the permanganate ion, \(MnO_4^-\) liberate \(O_2\) from water in the presence of an acid ?
Find the emf of the cell in which the following reaction takes place at 298 K
\(Ni(s) + 2Ag^+ (0.001\ M) \to Ni^{2+} (0.001\ M) + 2Ag(s)\)
(Given that \(E^\circ_{cell} = 10.5\) V, \(\frac{2.303\,RT}{F} = 0.059\) at 298 K)
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below :
BrO\(_4^-\) \(\xrightarrow{1.82\ V}\) BrO\(_3^-\) \(\xrightarrow{1.5\ V}\) HBrO \(\xrightarrow{1.595\ V}\) Br\(_2\) \(\xrightarrow{1.0652\ V}\) Br\(^-\)
Then the species undergoing disproportionation is
In the electrochemical cell :
\(Zn|ZnSO_4(0.01\ M)||CuSO_4(1.0\ M)|Cu\), the emf of this Daniel 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, \(\frac{RT}{F} = 0.059\))
The pressure of \(H_2\) required to make the potential of \(H_2\) - electrode zero in pure water at 298 K is :
