Question

help with these please

LU 101 is question. Consider the following half-reactions: JE° (V) Half-reaction Hg2+ (aq) + 2e → Hg(0.855V Ca2+(aq) + 2e —— C 0.403V Fe2+ (aq) + 2e -> Fe(s) 0.440V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Fe2+(aq) oxidize Hg() to Hg2+ (aq)? O y (6) Which species can be oxidized by Cd²+ (aq)? If none, leave box blank.

Answer 1

More negative the reduction potential, more the strength of reducing power.

Greater the reduction potential, greater the Oxidizing power.

Hence

1. strong oxidizing agent= Hg+2

2. Weak oxidizing agent: Cd+2( least reduction potential)

3. Weakest reducing agent: Hg

4. Strong reducing agent: Fe

5. No

6. Fe

2. Cell with more reduction potential acts as cathode and the other as anode.

Reaction at anode: Mn(s)------> Mn+2(aq)+2e-, Would=1.180 V

Reaction at cathode: 2 Cu+2(aq) +2e- ------> 2Cu+(aq)

Net Reaction: Mn(s)+2Cu+2(aq)---> Mn+2(aq)+ 2Cu+(aq)

Eocell= Eoanode+ Eocathode

= 1.180+0.153

Eocell=1.233 V , positive ,sspontaneous reaction.

3. Reaction at cathode: Co+2(aq)+2e- -----> Co(s), EO=-0.280 V

Reaction at anode: Mg(s) ----> Mg+2(aq) +2e-, EO= 2.370 V

Net Reaction: Mg(s) + Co+2(aq) -----> Co(s)+ Mg+2(aq)

Eocell= 2.370+(-0.280)

=2.370-0.280

Eocell= 2.090 V, positive, spontaneous