Question 9 of 15 Available From Not Set Due Date: Points Possibl 100 Grade Categ
ID: 102863 • Letter: Q
Question
Question 9 of 15 Available From Not Set Due Date: Points Possibl 100 Grade Category: Homework Map 12/8/20171 Sapling Learning Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2+1-0796 M and [Ni21-0.0130 M. Standard reduction potentials can be found here. Mg(s)-Ni2+ (aq) Mg2+ (aq)+ Ni(s) Policies: Homework Number can check your a You nswers. You can view solutions when yo give up on any question. You can keep trying to answer until you get it right or give up You lose 5% of the points availa answer in your question for ea attempt at that answer eTextbook Help With This Topic D Web Help & Videos Previous Give Up & View Solution O Check Answer Next Exit Technical and HintExplanation / Answer
When the cell is NOT under standard conditions, i.e. 1M of each reactants at T = 25°C and P = 1 atm; then we must use Nernst Equation.
The equation relates E°cell, number of electrons transferred, charge of 1 mol of electron to Faraday and finally, the Quotient retio between products/reactants
The Nernst Equation:
Ecell = E0cell - (RT/nF) x lnQ
In which:
Ecell = non-standard value
E° or E0cell or E°cell or EMF = Standard EMF: standard cell potential
R is the gas constant (8.3145 J/mol-K)
T is the absolute temperature = 298 K
n is the number of moles of electrons transferred by the cell's reaction
F is Faraday's constant = 96485.337 C/mol or typically 96500 C/mol
Q is the reaction quotient, where
Q = [C]^c * [D]^d / [A]^a*[B]^b
pure solids and pure liquids are not included. Also note that if we use partial pressure (for gases)
Q = P-A^a / (P-B)^b
substitute in Nernst Equation:
Ecell = E° - (RT/nF) x lnQ
Ni2+ + 2 e Ni(s) 0.25; Mg2+ + 2 e Mg(s) 2.372
E° = Ecathode - Eanode = -0.25 - -2.372 = 2.122 V
Ecell = 2.122 - (8.314*298)/(2*96500) * ln ( [Mg+2]/[Ni+2])
Ecell = 2.122 - (8.314*298)/(2*96500) * ln ( 0.796/0.013 )
Ecell = 2.0691 V
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