A voltaic cell is constructed with an Ag/Ag+ half-cell and a Pb/Pb2+ half-cell. Measurement shows that thesilver electrode is positive.

-Write balanced half-reactions and the overall spontaneous reaction.

redox eq

Ag^+(aq)+e^- >>> Ag(s)
Oxidation Eq
Pb(s) >>> Pb^2+(aq)+2e^-
overall
2Ag^+(aq)+Pb(s) >>> Pb^2+(aq)+2Ag(s)

Pb ==> Pb^2+ + 2e

2Ag^+ + 2e ==> 2Ag
--------------------
Pb + 2Ag^+ ==> Pb^2+ + 2Ag
Pb(s)|Pb^2+||Ag^+|Ag(s)
Pb is -; Ag is +

Sure, let's break this down into half-reactions and the overall reaction, but we'll do it with a dash of humor!

1. Ag/Ag+ Half-Cell:
Ag → Ag+ + e-
Here, Mr. Silver (Ag) donates an electron like throwing confetti at a party, forming a silver ion (Ag+).

2. Pb/Pb2+ Half-Cell:
Pb + 2e- → Pb2+
Mr. Lead (Pb) accepts two electrons like catching a game-winning touchdown pass, transforming into a lead ion (Pb2+).

Now, let's mix these two half-reactions and see what kind of chemical comedy unfolds!

3. Balancing the Half-Reactions:
2Ag + Pb → 2Ag+ + Pb2+
Two fun-loving silvers (2Ag) join forces with Mr. Lead (Pb), resulting in two silver ions having a blast (2Ag+) and a lead ion (Pb2+).

4. Overall Spontaneous Reaction:
2Ag + Pb → 2Ag+ + Pb2+
This reaction showcases the comedic chemistry of two lively silvers (2Ag), who get all charged up and turn into two silver ions (2Ag+), while Mr. Lead (Pb) also makes an appearance as a lead ion (Pb2+).

So, the overall spontaneous reaction is 2Ag + Pb → 2Ag+ + Pb2+, leaving us with a chemical performance full of electrifying fun and a positive silver electrode to light up the chemical stage!

To write the balanced half-reactions and the overall spontaneous reaction for the voltaic cell constructed with the Ag/Ag+ half-cell and the Pb/Pb2+ half-cell, we need to consider the standard reduction potentials for each half-reaction.

First, let's determine the half-reaction for the Ag/Ag+ half-cell. The standard reduction potential for Ag+ (aq) + e- → Ag (s) is +0.80 V.

In the Pb/Pb2+ half-cell, the standard reduction potential for Pb2+ (aq) + 2e- → Pb (s) is -0.13 V.

Since the silver electrode is positive, it indicates that the reduction potential for the Ag/Ag+ half-cell is larger than the reduction potential for the Pb/Pb2+ half-cell. Therefore, the Ag/Ag+ half-reaction will be the reduction half-reaction, and the Pb/Pb2+ half-reaction will be the oxidation half-reaction.

Writing the balanced half-reactions:

Reduction half-reaction:
Ag+ (aq) + e- → Ag (s)

Oxidation half-reaction:
Pb (s) → Pb2+ (aq) + 2e-

To balance the reduction half-reaction, we need two electrons, so we multiply the oxidation half-reaction by 2:

2Pb (s) → 2Pb2+ (aq) + 4e-

Next, we multiply the reduction half-reaction by 4 to balance the number of electrons:

4Ag+ (aq) + 4e- → 4Ag (s)

Now, we can combine the individual half-reactions to obtain the overall spontaneous reaction:

4Ag+ (aq) + 4e- + 2Pb (s) → 4Ag (s) + 2Pb2+ (aq)

Simplifying the equation:
4Ag+ (aq) + 2Pb (s) → 4Ag (s) + 2Pb2+ (aq)

This is the balanced overall spontaneous reaction for the given voltaic cell.

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