Question

The standard reduction potential of Zn and Ag in water at 298 K are, Zn$^{2+}$ + 2e$^{-}$ $\rightleftharpoons$ Zn; E$\circ$ = -0.76 V and Ag$^{+}$ + e$^{-}$ $\rightleftharpoons$ Ag;E$\circ$= +0.80 V. Which of the following reactions take place?
a- Zn$^{+2}$(aq) +2Ag(s) $\rightarrow$ 2Ag$^{+}$ (aq) + Zn(s)
b- Zn (s) +2Ag$^{+}$ (aq) $\rightarrow$ Zn$^{2+}$ (aq) + 2Ag(s)
c- Zn$^{+2}$(aq) +Ag$^{+}$(aq) $\rightarrow$ Zn(s) + Ag(s)
d- Zn (s) + Ag(s) $\rightarrow$ Zn$^{2+}$ (aq) + Ag$^{+}$(aq)

Answer 1

Hint: We are given with the standard reduction potential of both the metals, i.e Zn, and Ag. Identify the reactions occurring at anode, and at the cathode on the basis of reduction potential. Then, the reaction could be known.

Complete step by step answer:
First, let us know about the standard reduction potential. It is considered to be the reduction potential of metals, or ions under the specific conditions. It is the difference between the potential difference of the reactions occurring at the anode, and cathode.
As we know, the specific conditions for standard reduction potential are 298K, 1atm, and 1M solutions.
Now, we can see in the question that the reduction potential of Ag is more than the Zn.
Thus, silver will try to reduce; it means the reaction of zinc will occur at the anode, and it will oxidise (lose electrons).
The reaction of zinc at anode is Zn $\rightarrow$ Zn$^{2+}$ + 2e$^{-}$
Now, the reaction of silver at cathode is Ag$^{+}$ + e$^{-}$ $\rightarrow$ Ag(s), But the silver will gain two electrons, then the reaction will be 2Ag$^{+}$ + 2e$^{-}$ $\rightarrow$ 2Ag(s)
Now, to get the cell reaction we will combine the reaction occurring at anode, and cathode.
Then, the cell reaction is Zn (s) +2Ag$^{+}$ (aq) $\rightarrow$ Zn$^{2+}$ (aq) + 2Ag(s).
Hence, the correct option is (B).

Note: Don’t get confused between the words oxidising and reducing. Just remember that the molecule losing electrons will oxidise, and gain of electrons by the molecule means it will reduce. The above mentioned reactions will help to understand this concept.