7. You set up a voltaic cell: Sn(s) | Sn2+ (a)Fe (aq) , Fe (aq) Pt(s) Each half-

Question

7. You set up a voltaic cell: Sn(s) | Sn2+ (a)Fe (aq) , Fe (aq) Pt(s) Each half-cell contains 1.00 L of solution containing the appropriate aqueous speci concentrations. es in 1.00 M Initially, the tin electrode weighs 205 g and the platinum electrode weighs 100 g. You connect the electrodes with a wire and allow electrons to flow in the spontaneous direction. You measure a current of 0.250 A for 24.0 hours. After this time, you weigh the tin and platinum electrodes. What will be their weights? What will be the final value of [Sn2+] on the anode side of the cell?

Explanation / Answer

According to first law of electrolysis, amount of any substance deposited or dissolved at any electrode is directly proportional to the quantity of electricity passed through the system and is given by:

W = Z x I x T,

where W = amount of substance deposited or dissolved

Z= E/96500, E= equivalent weight

I= Current passed in ampere

T= times in sec

Given, I = 0.250 A, and T = 24 h = 86400 sec

--> For Sn electrode,

E = Molar mass/ Number of electrons involved = 118.7/2 =59.35 gm

Therefore, W = (59.35/96500) x 0.250 x 86400 = 13.28 gm of Sn is dissolved to the solution since it is the anode in the form of Sn2+.

Therefore, new weight of Sn electrode = 205-13.28 = 191.72 gm

--> For Fe3+/Fe2+ reaction on Pt electrode,

E = Molar mass/ Number of electrons involved = 55.845/1 =55.845 gm

Therefore, W = (55.845/96500) x 0.250 x 86400 = 12.5 gm of Fe is deposited on the Pt electrode since it is the anode in the form of Fe2+.

Therefore, new weight of Pt electrode = 100+12.5 = 112.5 gm

Since, 13.28 gm of Sn2+ is dissolved in the solution of volume=1 L,

Therefore, [Sn2+] = (Given Mass/Molar mass)/ Volume = 0.11 M

Related Questions

Navigate

• Browse (All)
• Browse 7
• Subjects

---

---

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