Pb(s) + Cu^(2+) (aq) = Pb(2+) (aq) + Cu (s). Suggest a change to one half-cell to increase the cell potential Answer question

Question

Pb(s) + Cu^(2+) (aq) = Pb(2+) (aq) + Cu (s). Suggest a change to one half-cell to increase the cell potential Answer question
🧐 Not the exact question you are looking for?Go ask a question

Solution 1

To increase the cell potential, you could increase the concentration of Cu^2+ ions in the solution. According to the Nernst equation, an increase in the concentration of the ions in the solution will increase the cell potential. This is because the reaction will be more likely to proceed in the dire Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study prob

Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solve study problem.
Knowee AI StudyGPT is a powerful AI-powered study tool designed to help you to solv

This problem has been solved

Similar Questions

Pb(s) + Cu^(2+) (aq) = Pb(2+) (aq) + Cu (s). Suggest a change to one half-cell to increase the cell potential Answer question

In a zinc-copper cell, Zn | Zn(1M) || Cu(1M) | Cu, which electrode is negative?Group of answer choicesCu(s)Zn(s)Zn(aq)Cu(aq)

In the following chemical reaction, which substance is the oxidizing agent and which is the reducing agent?Pb(s)  +  2 AgNO3(aq)  →  Pb(NO3)2(aq)  +  2 Ag(s)

alculate the cell voltage for a Daniell cell Zn(s) + Cu2+(aq) 🡪 Zn2+(aq) + Cu(s) at 50 °C, Zn2+ concentration: 10-3 M & Cu2+ concentration: 10-1 M

2Cu+ (aq) → Cu (s) + Cu+ (aq) 2Cu+ (𝑎𝑞) → Cu (𝑠) + Cu+ (𝑎𝑞) is +0.36 V at 298 K. The equilibrium constant of the reaction is

1/3