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e Economics question lCh x E Homepage ECON1B0. × -Bonus Pre Test Assignm x @Take a Test-Aashraya l /@Quizzes- HEMIE03 e x Secure https:/ avenue c mcnasier.ca d2 ms quizzing user/attempt quiz start ramed2l?ou=213383&isprv;=&drc;=0&q; 1692&c; =08 dnb-o ooo 4 ·-ile avenue to learn CHEM 1E03:General Chemistry for Engineering I ashraya Mehta Content Resources Communication AssessmentsPebblePad Quiz 6 St. Length: 166:39:00 ya Mehla: Allempl 1 Question 3 (1 point) 0 of 10 questions saved A voltaic cell at 25 C consists of Mn/Mn2+ and Cd/Cd2+ half-cells with the initial concentrations [Mn21-0.100 M and [Cd2+]-0.0100 M. Use the Nernst equation to calculate E for this cell. Page 1: Mn+2(aq) + 2e Mn(s) Eo=-1.18 V +0.78 V 8 +0.75 V 10 +0.72 V +1.61 Quiz Started +0.81 V O Type here to search 2017-11-25

Explanation / Answer

3)

from data table:

Eo(Mn2+/Mn(s)) = -1.18 V

Eo(Cd2+/Cd(s)) = -0.40 V

the electrode with the greater Eo value will be reduced and it will be cathode

here:

cathode is (Cd2+/Cd(s))

anode is (Mn2+/Mn(s))

The chemical reaction taking place is

Cd2+(aq) + Mn(s) --> Cd(s) + Mn2+(aq)

Eocell = Eocathode - Eoanode

= (-0.40) - (-1.18)

= 0.78 V

Number of electron being transferred in balanced reaction is 2

So, n = 2

we have below equation to be used:

E = Eo - (2.303*RT/nF) log {[Mn2+]^1/[Cd2+]^1}

Here:

2.303*R*T/F

= 2.303*8.314*298.0/96500

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[Mn2+]^1/[Cd2+]^1}

E = 0.78 - (0.0591/2) log (0.1^1/0.01^1)

E = 0.78-(2.956*10^-2)

E = 0.75 V

Answer: + 0.75 V

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know

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