Question

Calculate the standard cell emf for the reaction: ClO3−(aq)+3Cu(s)+6H+(aq)→Cl−(aq)+3Cu2+(aq)+3H2O(l) Pt is used as an inert electrode in contact with the ClO3− and Cl−. Calculate the standard emf using data in Appendix E and given the following: ClO3−(aq)+6H+(aq)+6e−→Cl−(aq)+3H2O(l); E∘=1.45 V Express the emf to three significant figures with the appropriate units. V

Answer 1

Answer: The standard EMF of the cell is 1.11 V

Explanation:

For the given chemical equation:

$ClO_3^-(aq)+3Cu(s)+6H^+(aq.)\rightarrow Cl^-(aq.)+3Cu^{2+}(aq.)+3H_2O(l)$

The half reaction follows:

Oxidation half reaction:  $Cu(s)\rightarrow Cu^{2+}+2e^-;E^o_{Cu^{2+}/Cu}=0.34V$    ( × 3)

Reduction half reaction: $ClO_3^-(aq.)+6H^+(aq.)+6e^-\rightarrow Cl^-(aq.)+3H_2O(l);E^o=1.45V$

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

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

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

Calculating the $E^o_{cell}$ using above equation, we get:

$E^o_{cell}=1.45-0.34=1.11V$

Hence, the standard EMF of the cell is 1.11 V

Answer 2

The standard cell potential is 1.11 V,

What is the standard cell emf?

The standard cell emf is obtained by the use of the formula;

E°cell = E°cathode - E°anode

Given that the reduction half equation is;  ClO3−(aq)+6H+(aq)+6e−→Cl−(aq)+3H2O(l); E∘=1.45 V and the oxidation half equation is; Cu(s) ---> Cu^2+(aq) + 2e E∘=0.34  V

Now; the standard cell potential is easily obtained from;

1.45 V - 0.34  V = 1.11 V

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