Question

Calculate the value of E°cell for the following reaction:2Au(s) + 3Ca2+(aq) → 2Au3+(aq) + 3Ca(s)Au3+(aq) + 3e- → Au(s) E° = 1.50VCa2+(aq) + 2e- → Ca(s) E° = -2.87VA) -4.37 VB) -1.37 VC) -11.6 VD) 1.37 VE) 4.37 V

Answer 1

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

Explanation:

For the given cell reaction:

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

The half reactions 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 )

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

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

Putting values in above equation, we get:

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

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

Answer 2

The  E°cell is calculated from the data provided to be  4.37 V.

An electrochemical cell is a cell that produces energy by spontaneous chemical processes. The anode is the positive electrode while the cathode is the negative electrode.

The E°cell is obtained from;

E°cell = E°cathode - E°anode

E°cell = 1.50V - (-2.87V) = 4.37 V

The  E°cell is calculated from the data provided to be  4.37 V.

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