Question

Given the half-reactions: $Cu^{2+}(aq) + 2e^- \to Cu(s)$ $E^0 = 0.34 V$ $Fe^{2+}(aq) + 2e^- \to Fe(s)$ $E^0 = -0.44 V$ If a voltaic cell using these two half-reactions and using "active electrodes" is formed, Question 1 (1) What is the reaction happened in the cathode? $Cu^{2+}(aq) + 2e^- \to Cu(s)$ $Fe^{2+}(aq) + 2e^- \to Fe(s)$ $Cu(s) \to Cu^{2+}(aq) + 2e^-$ $Fe(s) \to Fe^{2+}(aq) + 2e^-$

Name: given the half reactions cu2aq 2e to cus e0 034 v fe2aq 2e to fes e0 044 v if a voltaic cell using these two half reactions and using active electrodes is formed question 1 1 what is the rea 63476
Uploaded: 2022-04-26T06:33:52-08:00
Duration: 2 min 43 s
Channel: Shaiju T
Description: given the half reactions cu2aq 2e to cus e0 034 v fe2aq 2e to fes e0 044 v if a voltaic cell using these two half reactions and using active electrodes is formed question 1 1 what is the rea 63476

          Given the half-reactions:
$Cu^{2+}(aq) + 2e^- \to Cu(s)$  $E^0 = 0.34 V$
$Fe^{2+}(aq) + 2e^- \to Fe(s)$  $E^0 = -0.44 V$
If a voltaic cell using these two half-reactions and using "active electrodes" is formed,

Question 1
(1) What is the reaction happened in the cathode?
$Cu^{2+}(aq) + 2e^- \to Cu(s)$
$Fe^{2+}(aq) + 2e^- \to Fe(s)$
$Cu(s) \to Cu^{2+}(aq) + 2e^-$
$Fe(s) \to Fe^{2+}(aq) + 2e^-$

given the half reactions cu2aq 2e to cus e0 034 v fe2aq 2e to fes e0 044 v if a voltaic cell using these two half reactions and using active electrodes is formed question 1 1 what is the rea 63476

Added by Melanie A.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Shaiju T

04/26/2022

Step 1

Reduction is the gain of electrons. We need to find the half-reaction with the more positive reduction potential, as this will be the reaction that occurs at the cathode in a voltaic cell. Show more…

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Given the half-reactions: $Cu^{2+}(aq) + 2e^- \to Cu(s)$ $E^0 = 0.34 V$ $Fe^{2+}(aq) + 2e^- \to Fe(s)$ $E^0 = -0.44 V$ If a voltaic cell using these two half-reactions and using "active electrodes" is formed, Question 1 (1) What is the reaction happened in the cathode? $Cu^{2+}(aq) + 2e^- \to Cu(s)$ $Fe^{2+}(aq) + 2e^- \to Fe(s)$ $Cu(s) \to Cu^{2+}(aq) + 2e^-$ $Fe(s) \to Fe^{2+}(aq) + 2e^-$