My bench average was 0.178v Cathode Compartment Anode Compartment Ag(s) Ag (0.05

Question

My bench average was 0.178v

Cathode Compartment Anode Compartment Ag(s) Ag (0.0500 M) Ag(s) Ag (x M) (21) o 0.059 (14) Using the Nernst Equation og10Q cel Where (Delly 0.00 v, n the number of electron needed to reduce Ag ions Ag (aq) e Ag(s) E Ag 0.059 [Ag product Equation (14) become equation (22) cell n log 0.0500 (22) and Q IAg lreactant Step 1 the [Ag can be obtained from equation (22) 0.0591. [Ag log where E is your experimentally measured voltage 11 0.0500 Step 2 the CC n can be determined from the dilution equation M1v1 E M2V2 (23) [cn spenced. see step 21 of mixture) Step 3 once the [Ag and [cnhas been obtained it is a simple matter of plugging these values into the Ksp expression for silver chloride

Explanation / Answer

Given;

Ecell = 0.178 V

Ecell = - .0591/1 log [ Ag+(xM)] / .05

Ag+(aq) + Cl -(aq) -----> AgCl(s)

The reaction is of strong acid and strong base.

so,

[Ag+] = xM [Cl-] = xM

X^2= Ksp

.0178 = .0591/1 log [xM] / .0500

0.2371 = log[xM] /.0500

x = 0.06338 M

Ksp = (.06338 )^2

Ksp = 0.0040 M

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