Calculate the value of E°cell for the following reaction:2Au(s) + 3Ca2+(aq) → 2Au3+(aq) + 3Ca(s)Au3+(aq) + 3e- → Au(s) E° = 1.50

Question

20 days ago

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VCa2+(aq) + 2e- → Ca(s) E° = -2.87VA) -4.37 VB) -1.37 VC) -11.6 VD) 1.37 VE) 4.37 V

1 answer:

Alekssandra [2.6K] · Answer 1 · 2026-03-16T00:40:39+03:00

Answer: The standard electrode potential of the cell is -4.37 V

Explanation:

For the specified cell reaction:

$2Au(s)+3Ca^{2+}(aq.)\rightarrow 2Au^{3+}(aq.)+3Ca(s)$

The half reactions are as follows:

Oxidation half reaction: $Au(s)\rightarrow Au^{3+}(aq.)+3e^-;E^o_{Au^{3+}/Au}=1.50V$ ( × 2 )

Reduction half reaction: $Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V$ ( × 3 )

In this scenario, the substance being oxidized acts as the anode, while the one being reduced becomes the cathode.

To determine the $E^o_{cell}$ of the reaction, we utilize the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

By substituting the appropriate values into the above equation, we arrive at:

$E^o_{cell}=-2.87-(1.50)=-4.37V$

Thus, the standard electrode potential of the cell is -4.37 V