The standard reduction potentials for two half-reactions at 25^\circ\text{C} are given bel… — JAMB UTME

The standard reduction potentials for two half-reactions at

25^\circ\text{C}

are given below:

\text{Co}^{2+}(aq) + 2e^- \rightarrow \text{Co}(s) \quad E^\circ = -0.28\text{ V}

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s) \quad E^\circ = +0.34\text{ V}

What is the standard electromotive force ( $E^\circ_{\text{cell}}$ ) for the spontaneous reaction that occurs when these two half-cells are connected under standard conditions?

$+0.62\text{ V}$ Answer
B
$+0.06\text{ V}$
C
$-0.62\text{ V}$
D
$-0.06\text{ V}$

Answer

The standard electromotive force for the spontaneous cell reaction is

+0.62\text{ V}

For a spontaneous electrochemical cell, reduction occurs at the electrode with the more positive reduction potential (cathode:

\text{Cu}^{2+}/\text{Cu}

+0.34\text{ V}

), and oxidation occurs at the electrode with the more negative reduction potential (anode:

\text{Co}^{2+}/\text{Co}

-0.28\text{ V}

). Subtracting the anode potential from the cathode potential yields

E^\circ_{\text{cell}} = +0.34\text{ V} - (-0.28\text{ V}) = +0.62\text{ V}

Step-by-Step Solution

Identify the cathode and anode based on standard reduction potentials.

Copper half-cell (

E^\circ = +0.34\text{ V}

) has the higher reduction potential and acts as the cathode (reduction). Cobalt half-cell (

E^\circ = -0.28\text{ V}

) has the lower reduction potential and acts as the anode (oxidation).

A species with a higher reduction potential is more easily reduced, driving spontaneous electron flow from anode to cathode.

Calculate the standard cell electromotive force (

E^\circ_{\text{cell}}

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.34\text{ V} - (-0.28\text{ V}) = +0.34\text{ V} + 0.28\text{ V} = +0.62\text{ V}

The cell EMF for a spontaneous reaction must be positive.

Key Concept

Standard Cell Potential and Reaction Spontaneity

Estimated Time:1m 0s

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