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defon
3 years ago
14

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

VCa2+(aq) + 2e- → Ca(s) E° = -2.87VA) -4.37 VB) -1.37 VC) -11.6 VD) 1.37 VE) 4.37 V
Chemistry
1 answer:
uranmaximum [27]3 years ago
8 0

<u>Answer:</u> The standard electrode potential of the cell is -4.37 V

<u>Explanation:</u>

For the given cell reaction:

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

The half reactions follows:

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

<u>Reduction half reaction:</u>  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

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