The correct value of cell potential in volts for the reaction that occurs when the following two half cells ar

Question

The correct value of cell potential in volts for the reaction that occurs when the following two half cells are connected, is:
$Fe^{2+}(aq) + 2e^- ightarrow Fe(s) \ ; \ E^{\circ} = -0.44	ext{ V}$
$Cr_2O_7^{2-}(aq) + 14H^+ + 6e^- ightarrow 2Cr^{3+} + 7H_2O \ ; \ E^{\circ} = +1.33	ext{ V}$

Accepted Answer

To determine the correct cell potential ($E^{\circ}_{cell}$), we identify the cathode and the anode based on their standard reduction potentials ($E^{\circ}$).
The half-cell with the higher standard reduction potential will undergo reduction and act as the cathode. The half-cell with the lower standard reduction potential will undergo oxidation and act as the anode.
Given:
$E^{\circ}_{Cr_2O_7^{2-}/Cr^{3+}} = +1.33	ext{ V}$ (Cathode)
$E^{\circ}_{Fe^{2+}/Fe} = -0.44	ext{ V}$ (Anode)

The standard cell potential is calculated as:
$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$
$E^{\circ}_{cell} = 1.33	ext{ V} - (-0.44	ext{ V})$
$E^{\circ}_{cell} = 1.33 + 0.44 = +1.77	ext{ V}$

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