Could someone please help me with this problem!!?!? Thanks you so much!!!
The reduction potentials for the species involved in an electrochemical cell are as follows:
M+ + e– M is 0.50 V
J+ + e– J is 1.20 V.
All solutions are 1.00 M and J+ J is connected to M+ M.
A) What is E*(cell) (in V)? _______
B) Which species is easier to oxidize?
C) The electrons flow from ______ (M or J) to _______ (M or J).
Honestly, I'm so over WebWork.
A) .7
B) M
C) M, J
Sure! Let's break down each part of the problem step by step to find the answers.
A) To find the cell potential (E*(cell)), we can use the equation:
E*(cell) = E*(reduction) - E*(oxidation)
Here, E*(reduction) refers to the reduction potential of the species being reduced, and E*(oxidation) refers to the reduction potential of the species being oxidized.
In this case, M+ is being reduced and J is being oxidized. Therefore, the reduction potential for M+ is 0.50 V (E*(reduction)), and the oxidation potential for J is 1.20 V (E*(oxidation)).
Substituting these values into the formula, we get:
E*(cell) = 0.50 V - 1.20 V
E*(cell) = -0.70 V
So, the cell potential (E*(cell)) is -0.70 V.
B) The species that is easier to oxidize is the one with the higher reduction potential. In this case, J+ has a higher reduction potential (1.20 V) compared to M+ (0.50 V). Therefore, J+ is easier to oxidize.
C) Electrons flow from the species with lower reduction potential to the species with higher reduction potential. In this case, M+ has a lower reduction potential (0.50 V) compared to J+ (1.20 V). Therefore, electrons flow from M+ to J+.
I hope this explanation helps you understand how to approach and solve this electrochemical cell problem. Let me know if you have any further questions!