Solveeit Logo
Login

Question

Question: \(\text{CuI(s) + }{{\text{e}}^{\text{-}}}\text{ }\to \text{ Cu(s) + }{{\text{I}}^{\text{-}}}\); \({{...

CuI(s) + e- → Cu(s) + I-\text{CuI(s) + }{{\text{e}}^{\text{-}}}\text{ }\to \text{ Cu(s) + }{{\text{I}}^{\text{-}}}; E0Cu+/Cu{{E}^{0}}_{C{{u}^{+}}/Cu} = -0.16 V
Zn2+(aq) + 2e- → Zn(s)\text{Z}{{\text{n}}^{\text{2+}}}\text{(aq) + 2}{{\text{e}}^{\text{-}}}\text{ }\to \text{ Zn(s)} ; E0Zn2+/Zn{{E}^{0}}_{Z{{n}^{2+}}/Zn} = -0.76 V
Calculate the EMF:
(A) 0.90 V
(B) 0.45 V
(C) 0.60 V
(D) 0.75 V

Explanation

Solution

Identify the oxidation half reaction and reduction half reaction. Based on that we can arrange the electrode potential values to find the EMF. It is generally considered that oxidation happens at anode and reduction happens at cathode. The formula for EMF thus becomes:
EMF = E0cathode−E0anode{{E}^{0}}_{cathode}-{{E}^{0}}_{anode}
Where, the potential values are taken in standard electrode potential.

Complete step-by-step answer:
Electrochemistry is the branch of physical chemistry that mainly deals with the relationship between electricity and identifiable chemical change.
Electrochemical reaction is a chemical reaction in which current is externally supplied or produced through a spontaneous chemical reaction.
Chemical reactions where electrons are directly transferred between the constituent molecules or atoms are called oxidation-reduction or rather redox reactions.
We have to identify the cell reaction that can possibly take place when electricity is passed through the circuit. For this we need to refer to the standard electrode potential values of the two half-cell reactions.
The oxidation of zinc(E0Zn2+/Zn{{E}^{0}}_{Z{{n}^{2+}}/Zn} = 0.76 V ) provides more potential value than the corresponding oxidation of copper metal(E0Cu+/Cu{{E}^{0}}_{C{{u}^{+}}/Cu} = 0.16 V).
The oxidation of zinc takes place at the positive electrode i.e. anode. The reduction of copper takes place at the negative electrode i.e. cathode.
Thus, the cell reaction becomes :
Zn + 2Cu+→ Zn2+ + 2Cu\text{Zn + 2C}{{\text{u}}^{\text{+}}}\to \text{ Z}{{\text{n}}^{\text{2+}}}\text{ + 2Cu}
We will now calculate the EMF of the cell by using the formula mentioned in the hint.
EMF = E0cathode−E0anode{{E}^{0}}_{cathode}-{{E}^{0}}_{anode}
EMF = ( -0.16) - (- 0.76) = 0.60 V
Therefore, the EMF of the cell is 0.60 V and the correct answer is option (C).

Note: A salt bridge is used in chemical reactions where the anode solution is different from a cathode solution. In this case salt bridge is used to complete the circuit so that there is passage of current and hence chemical reaction takes place.