When the Pb²+ concentration is 5.44x104 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 0.953V. What is the Mn²+ concentration? Pb²+ (aq) + Mn(s)→→→→→ Pb(s) + Mn²+(aq) Answer: M

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.98QE
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When the Pb²+ concentration is 5.44×104 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 0.953V. What is the Mn²+
concentration?
Pb²+ (aq) + Mn(s) Pb(s) + Mn²¹(aq)
Answer:
M
Transcribed Image Text:When the Pb²+ concentration is 5.44×104 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 0.953V. What is the Mn²+ concentration? Pb²+ (aq) + Mn(s) Pb(s) + Mn²¹(aq) Answer: M
When the
Pb²+ concentration is 1.11 M, the observed cell potential at 298 K for an electrochemical cell with the following reaction is 0.676 V. What is the
Cr³+ concentration?
3Pb²+ (aq) + 2Cr(s) → 3Pb(s) + 2Cr³+ (aq)
red
Pb²+ (aq) + 2e¯ → Pb(s) Ee
→ Cr(s) E red
Cr³+ (aq) + 3e
[Cr³+ ]
=
M
-0.126 V
= -0.740 V
Transcribed Image Text:When the Pb²+ concentration is 1.11 M, the observed cell potential at 298 K for an electrochemical cell with the following reaction is 0.676 V. What is the Cr³+ concentration? 3Pb²+ (aq) + 2Cr(s) → 3Pb(s) + 2Cr³+ (aq) red Pb²+ (aq) + 2e¯ → Pb(s) Ee → Cr(s) E red Cr³+ (aq) + 3e [Cr³+ ] = M -0.126 V = -0.740 V
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