Question

Z. Consider the following data: E(Ag O/Ag- +1.36 V; EAg g")-+1.98 V; E(Ag/Ag)-+0.80 V; E (Ag.S/AgH2S)--0.69 v (a) Which of the five species (Ag S, Ag:O, Ag g,Ag) is the strongest oxidizer and which one should be the easiest to oxidize? (b) The standard hydrogen electrode is connected with an electrode made of silver and containing Ag ions in water as shown below: Ag Which of the electrodes is the cathode and which is the anode? Which way the electrons should move through the wire? Write the balanced chemical reaction taking place in the battery. (c) Decide whether silver can be oxidized by HS under the standard conditions. If yes, write the balanced reaction equation and calculate the e (d) Which of the five silver species is most likely to be reduced by water (Note: E(O2H'/H20)1.23 V)? Write the balanced reaction equation. (e) Write a balanced half-reaction for E(Ag oVAg 8. Consider the disproportionation of Ag. Complete and balance this reaction and predict the sign of the free energy change based on the Ecl

Answer 1

Solution:

a) The value of electrode potential is highest for E⁰(Ag²⁺ / Ag⁺) = +1.98 V.

Hence, Ag²⁺ is the strongest oxidizing agent.

Since, the value of electrode potential is lowest for E⁰(Ag₂S/  Ag, H₂S) = -0.69 V.

Hence, Ag₂S is the strongest reducing agent and can easily oxidize.

b) Since, E⁰(Ag⁺ / Ag) = +0.80 V and E⁰ for H₂/H⁺ electrode is 0.00 V.

When the half cell is combined with H₂/H+ half cell, then Ag⁺/ Ag is working as cathode, where oxidation occurs

and H₂/H+ works as anode also called reduction half cell.

Then cell is written as, E⁰(Ag⁺ / Ag) || E⁰(H⁺ / 1/2 H₂)

Cell reactions are ;

At cathode, Ag  ===== Ag⁺ + e-

At anode, H⁺ + e- ======1/2 H2

Electrons are generated at cathode and moves towards anode.

C) Yes, Ag can be oxidized by H₂S

Since, Ag₂S === 2Ag + H₂S, E⁰ = - 0.69,

Therefore, silver can be oxidized into Ag₂S

The reaction will be,

2Ag + H₂S === Ag2S + H₂, E⁰ = + 0.69 V,

E⁰cell = +0.69 V

D) Since, the highest positive emf for;

E⁰(Ag²⁺ / Ag⁺) = +1.98 V.

Hence, Ag²⁺ is easily reduced by water.

E) The balanced cell reaction for, Ag₂O₃ / Ag2+ wll be

Ag₂O₃ + 4 e- === 2Ag²⁺ + 3/2 O₂