Question

Can you help explaim and solve how to find delta G along with anything related to half-cells I dont even know where to begin, thank you!

WILL probe to the piece of metal foil • Read and record the voltape. Do not forget the NOTE : The The filter paper string should be placed in the welis just prior to measuring the voltage 10. Reverse the probes Record what happens to the sign and switch the meter off NOTE: The voltage (potential that recorded first is the electrochemical cell voltage of a cell, which is made up of two half-cells joined electrically by wires the probes and the salt bridge solution (NO) The red probe is the positive terminal and the black probe is the negative terminal. The probes are touched to the electrodes in the cell (Cu metal and Sn foil) and the measured voltage will have a positive sign of the black probe is on the Anode and the red robe is on the Cathode If the sign of the measured voltage is negative, then the reverse of the above is true the black probe is on the Cathode and the red probe is on the Anode REMEMBER The Anode is the electrode at which oxidation occurs. The Cathode is the electrode at which reduction occurs. The voltage read in step 9 should be positive. Since the black probe is on the Sn this must be the anode. Results: AG calculated) K(calculated) Shorthand Cell Designation Curss/cm2" || sn 241 sn E' (measured) 4.52 -.52. Show calculations for AGⓇ and K

Answer 1

Using your given information

For the cell

Cu/Cu²⁺// Sn²⁺/Sn

Left side is anode and right side is cathode.

Oxidation reaction (anode)

Cu - 2e $\rightarrow$ Cu²⁺.

Reduction reaction ( cathode)

Sn²⁺ +2e $\rightarrow$ Sn

Given, E^o_cell = E^o_cathode - E^o_anode = - 0.52 V.

Given E⁰_anode = E^o(Cu²⁺​​​​​​/Cu) = 0.34 V

E⁰_cathode = - 0.52 +0.34 = - 0.18 V.

$\Delta$G^o = - nFE^o_cell

n is number of electrons transfered .

For the above cell n = 2

F = 96500 C

$\Delta$G^o = - 2*96500*(-0.52) = 100360 J.

$\Delta$G^o = - RTlnK

Or, K = e^{-
G T
/RT}

R = 8.314 J/mol.K

^{at 298 K}

Or, K = e^{-(100360/8.314×298)}

Or, K = 2.56 ×10^-18.

But , for spontaneous reaction E⁰_cell must be positive. Therefore Sn must be at anode.

So, correct shorthand cell designation will be

Sn(s)/Sn²⁺// Cu⁺/Cu(s)

Then

Oxidation reaction (anode)

Sn - 2e $\to$ Sn²⁺

reduction reaction (cathode)

Cu²⁺ +2e $\to$ Cu

When , Sn is anode, E^o_cell = + 0.52 V.

Then, ^o = - 100360 J.

G T

and , K = e^{(100360/8.314*298)} = 3.90*10¹⁷