Question

Enter electrons as e^-.

A voltaic cell is constructed in which the anode is a Sn|Sn²⁺ half cell and the cathode is a I^-|I₂ half cell. The half-cell compartments are connected by a salt bridge.

(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)

The anode reaction is:

__(aq)(s)(l)(g)

+

__(aq)(s)(l)(g)

----->

__(aq)(s)(l)(g)

+

__(aq)(s)(l)(g)

The cathode reaction is:

__(aq)(s)(l)(g)

+

__(aq)(s)(l)(g)

----->

__(aq)(s)(l)(g)

+

__(aq)(s)(l)(g)

The net cell reaction is:

__(aq)(s)(l)(g)

+

__(aq)(s)(l)(g)

----->

__(aq)(s)(l)(g)

+

__(aq)(s)(l)(g)

In the external circuit, electrons migrate _____fromto the I^-|I₂ electrode _____fromto the Sn|Sn²⁺ electrode.

In the salt bridge, anions migrate _____fromto the Sn|Sn²⁺ compartment _____fromto the I^-|I₂ compartment.

Answer 1

anode reaction --> must be loss of electrons, from the list shown, I is that species, will donate e-

Sn(s) ---> Sn2+(aq) + 2e-

cathode --> accepts electrons, it will get reduced

2I-(aq) + 2e- --> I2(s)

net ionic:

Sn(s) ---> Sn2+(aq) + 2e-

2I-(aq) + 2e- --> I2(s)

--------------------------------

Sn(s) + 2I-(aq) + 2e- --> I2(s) +  Sn2+(aq) + 2e-

d)

electrons migrate from I- towards Sn2+

e)

salt bridge msut counter balance charge via ions so

Sn2+ is decerasing, cations must go to anod

I- is decreasing as well, anions must go to the cathode