The value of E M^(+) / M o of four metals A, B, C and D are -1.2 V, 0.6 V, 0.85 V and -0.76 V, respectively. The order of tendency of their…

Question

The value of E M^(+) / M o of four metals A, B, C and D are -1.2 V, 0.6 V, 0.85 V and -0.76 V, respectively. The order of tendency of their unipositive cations getting deposited at the cathode during electrolysis of the respective salts is

MARKS Content Team · Accepted Answer

Correct answer: (C) C>B>D>A.. During electrolysis, the species having higher reduction potential is preferably reduced at the cathode. As per the data given in the question, the standard reduction potential of the metals are in the order- E C ∘ > E B ∘ > E D ∘ > E A ∘ Hence, C>B>D>A will be the preference of deposition of these metals on the cathode.

The value of $(E_{M^{+} / M}^{o})$ of four metals $A, B, C$ and $D$ are $- 1.2 V, 0.6 V, 0.85 V$ and $- 0.76 V$ , respectively. The order of tendency of their unipositive cations getting deposited at the cathode during electrolysis of the respective salts is

Solution

More Electrochemistry Questions

The value of E M + / M o of four metals A, B, C and D are -1.2 V, 0.6 V, 0.85 V and -0.76 V, respectively. The order of tendency of their unipositive cations getting deposited at the cathode during electrolysis of the respective salts is

Solution

More Electrochemistry Questions

The value of $(E_{M^{+} / M}^{o})$ of four metals $A, B, C$ and $D$ are $- 1.2 V, 0.6 V, 0.85 V$ and $- 0.76 V$ , respectively. The order of tendency of their unipositive cations getting deposited at the cathode during electrolysis of the respective salts is