Question

no need long calculation correct answer is fine

W ces important values Il needed for this question. When the Agt concentration is 3.90x10M the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.775V. What is the Mn? concentration? 2Ag (aq) + Mn(s)—-2 Ag(s) + Mn?" (aq) Answer: M Submit Answer Retry Entire Group 3 more group attempts remaining

Answer 1

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PART 1:

Given reaction:

2 Agtag) + Mn(s) + 2 Ag(s) + Mno

Given that,

Concentration of Ag⁺,

Ag+] = 3.9 x 10-4 M

Ecell = 1.775 V

Standard cell potential is given by:

Ecell = E Ag+ /Ag - EMn/Mn2+

Ecell 1 = 0.799 V - (-1.19 V)

:: Ecell = 1.989 V

From Nernst equation, we have,

0.0591 :. Ecell = Ecell- -log Mn2+1 [Ag+] /

0.0591 ::1.775 V = 1.989 V - - log Mn2+1 3.9 x 10-4 M)

log|Mn2+1 = 3.833

:. Mn2+1 = 6808.12 M

Please post one question at a time. Anyways answer for part 2: 1.166 x 10^-3 M
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