I had trouble with understanding why they put 0.57M on the denominator of Q ulti

Question

I had trouble with understanding why they put 0.57M on the denominator of Q ultimatly multiplying it by everything and not dividing?

6) You create a voltaic cell with each half-cell consisting of a copper electrode immersed in CrCl3. If Ecell = 0.0200 V and [CrCle] = 0.57 M at the anode, what is [CrCl3] at the anode?

This is a concentration cell with half-reactions

Cr(s)Cr3+ (aq)+3e E =0.73V

Cr3+ (aq)+3e Cr(s) E =0.74V

E cell =0

Recalling that reduction occurs at the cathode, we realize that [CrCl3]=0.57 M is the reactant concentration.

Ecell =0.0592/n log(Q)

10 3* 0.0300 / 0.0592 = (CrCl3) / 0.570

= 0.055M

Explanation / Answer

Cell reaction:

Cr3+ (aq., 0.57M) + Cr (s) ----------> Cr(s) + Cr3+ (aq., ?M)

Applying Nernst equation for the cell reaction,

Ecell = E0cell - (0.0591/n) * Log[Cr3+]/[Cr3+]

0.0200 = 0.000 - (0.0591/3)*Log[Cr3+]/(0.570)

Log[Cr3+]/(0.570) = - 1.02

[Cr3+] / 0.570 = 10-1.02

[Cr3+] = 0.0955 * 0.570

[Cr3+] = 0.0544 M

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.