a. Ag^+ (aq) + e^- ------------------> Ag(s) E0 = 0.80v
n = 1
Ecell = E0cell -0.0592/n logQ
= 0.80 -0.0592/1log1/[Ag^+]
= 0.80-0.0592log1/0.0436
= 0.80-0.0592*.13606 = 0.79v
b. Fe^3+ (aq) + e^- --------> Fe^2+ (aq) E0 = 0.77v
n = 1
Ecell = E0cell -0.0592/n logQ
= 0.77 -0.0592/1log[Fe^2+]/[Fe^3+]
= 0.77-0.0592log0.09/0.000534
= 0.77-0.0592*2.2267 = 0.64v
c. AgBr(s) + e^----------------------> Ag(s) + Br^- E0 = 0.0713v
n =1
Ecell = E0cell -0.0592/n logQ
= 0.0713 -0.0592/1log[Br^-]
= 0.0713-0.0592log0.037
= 0.0713-0.0592*-1.4317 = 0.156v >>>>answer
22-1 Calculate the electrode potentials of the following half-cells. Ag (0.0436M)|Ag a. b. Fe3 (5.34x10-4M), Fe2...
+ Given the following electrode potentials at 25°C Fe3+ e-- Fe2+ E° = 0.571 V 2e Fe(s) E° = -0.440 V Calculate the electrode potential for Fe3+ + 3 e- Fe(s) Fe2+ + Select one: a. -0.132 b. -0.036 c. 0.081 d.-0.211 e. 0.103
7.53 Given the following electrode potentials at 25°C, Fe3+ +e- = Fe2+ E' = 0.771 V Fe2+ + e = Fe(s) E = -0.440 V calculate the electrode potential for [Fe3+ + e = {Fe E; = ? for
Half-cell Potentials: Half Reaction: E value +0.80 V Agt + e → Ag Fe3+ + € → Fe2+ +0.77 v +0.34 V -0.13 V Cu2+ +2e → Cu Pb2+ + 2e - → Ib Ni2+ + 2e → Ni Cd2+ +2e → Cd -0.25 V -0.40 V Fe2+ + 2e → Fe -0.44 V Zn2+ + 2e → Zn -0.76 V Al3+ +3e → AI - 1.66 V Consider an electrochemical cell constructed from the following half cells, linked by...
Given the following electrode potentials at 25°C Fe3+ + e- → Fe2+ E° = 0.171 V Fe2+ + 2e-→ Fe(s) E° = -0.440 V Calculate the electrode potential for Fe3+ + 3 e- → Fe(s) Select one: a.-0.236 b. 0.081 c.-0.036 d. -0.211 e. 0.121
Calculate the equilibrium electrode potential for Fe3+/Fe2+ redox system, if the initial concentration of Fe2+ is of 0.1 mol/L and one third of Fe2+ is oxidized to Fe3+ at pH 1. (eo(Fe3+/Fe2+) = 0.77 V)
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A galvanic cell is prepared using the following two half-cells: (i) MnO4 (0.273 M), Mn2+ (0.167 M), and H+ (1.0 M), and (ii) Fe2+ (0.247 M) and Fe3+ (0.389 M). The standard reduction potentials for the two half cells are: Mnoa + 8H+ + 5e – Mn2+ + 4H20 E° = 1.507 V Fe3+ + e- Fe2+ E° = 0.770 V a) Calculate the voltage for this galvanic cell. b) Write the balanced equation for...
u. 1 an anode? N 3. Write the half-reaction at each electrode if cells containing following redox couples are connected a. Bry/Br and Fe3+/Fe2+ b. CdCl/Cd and AICIJAI ILI
Please combine the Mn electrode half reaction with the Ag
electrode half reaction and write the complete redox reaction using
date from table 8.1. Indicate which is the oxidizing and which is
the reducing agent.
b. Calculate the EMF when the reaction quotient (Q) =
]Mn2+]/[Ag+]^2 = 10^-5
Reducing agent Half-reaction E0 (volts) -2.93 2.87 Ca Na Mg Mn Zn Fe Ni Pb Ca Ca2++2e Na → Na++e- -2.36 Mn Mn++2e Zn Zn2++2e Fe → Fe2++ 2e- NiNi2++2e- -0.76 -0.47...
A voltaic cell consists of an Al/Al3+ half-cell and an Fe2+/Fe3+ half-cell. Calculate Ecell when conc. of Fe2+= 0.402 M, Fe3+= 0.073 M and Al3+= 0.239 M. Use the reduction potentials for Al3+ is -1.66 V and for Fe3+ is 0.77 V. How do you decide which is oxi, red.
Using std. electrode potentials listed, calculate the equilibrium constant at 25 ⁰C for the following reaction: O2 (g) + 4 H+ (aq) + 4 Fe2+ (aq) → 4 Fe3+ (aq) + 2 H2O (l) Fe3+ + e- = Fe2+ Ered= 0.77 O2 + 4H+ + 4e- = 2 H20 Ered= 1.23