Question

1)

Balance the following reaction under acidic conditions and calculate the cell potential in (V) at 298 K generated by the cell. Report your answer to the hundredths place.

Cr₂O₇^2-(aq) + I^-(aq) → Cr³⁺(aq) + I₂(s)

[Cr₂O₇^2-] = 2.0 M, [H⁺] = 1.0 M, [I^-] = 1.0 M, [Cr³⁺] = 1.0 × 10^-5 M

2) What is the value of n for the following reaction? Enter the whole number.

3Ni⁺(aq) + Cr(OH)₃(s) + 5OH^-(aq) → 3Ni(s) + CrO₄^2-(aq) + 4H₂O(l)

Answer 1

Solution 1) First we have to balance this redox reaction Cr202 + I-lag), erstlaq) + Iz(s) (aq) (220,2- + 6e- + 14H® -> 263+ + 7420 = Reduction halt { 21 → In + Re"}x} . Oxidation half Cr2022 +670 +14H0 - 26,3+ + I + TH₂ Balanced redox reaction C cell celluviisil on from 1 nolved) FL 298* Now from Nemst equation of electrochemical al Ecce - Ecu – 350 FT boy. Re lle & las be- Eau o ibém - Doogee log [ex2432 x [1-34 2:3IRT = 0.0591 at dol l a [ex20,29 [H1934 *[7]). [H203 = 1 as excess Eau = { bandh - Card) - ooo, ly, a Esve (1:31 - 0-53V) osoy ley Quinny FC 0-80v ocasos boy Q 7 -0 Q = (1890-51(1.036 (2.0) (1,094 (1) as 12 (5) Q = 10-10 or TQ = 500X20-1- Put this value of eq ② in ego I potential values does not depend on value of no we have from Standard Reduction 1 Table Ecath z +1.33v| where reduction i locuns) i Win Eerst, Gra87 -} . Eanode z +0.53 by o (where oxidation € 12, 21- . As standard takes place) reduction

OV 0.059 Ecev = 0.80V - D50 hoy (50 x 10") 2 0.80 v = 0.00385 lag., (5 % 10-19 2 0.80V - 0.00985 { 19.5 -11] 01/1Y 2 0.80 V - 0009 20.80 V + 0.101 V = 0.901V Yecue = 0901 v) Any 0 . We have to calculate the value of in' for the no. of e- Involved in overall C reaction. following reaction i,e 3Nit (aq) + Colonia (s) + 50H0 —> 3Ni (59 + er 0,2-laq) + (99) Lolita 4H201 i) . This given equation is a balanced redox reaction, 3Ni+ +30 +3N ( Reduction - half) Hence it Involved [123] An van uw