Why are the non-standard state cell potentials the same as the standard state cell potentials ? Please help me explain why this occurs.
The only way a non-standard state cell potential will be the same as the std. state potential is:
- If they have the same reference, i.e. Hydrogen
NOTE that if we tlak about "differences"
this is due to the reference values...
The fundamental difference in reduciton potential is always kept, the reference will only change with respect to the original SHE potential
Why are the non-standard state cell potentials the same as the standard state cell potentials ?...
1. Use the experimentally-determined reduction potentials above to compute cell potentials (voltages) for the following cells. Put these experimental values in the Ecell column. Then Calculate the standard cell potentials for these cells using standard reduction tables. Put these standard theoretical values in the cell column. Experimental Theoretical Ecell E cell Ni2+ + Fe <==> Fe2+ + Ni Ni2+ + Mg <==> Mg2+ + Ni Zn2+ + Ni <==> Zn + Ni2+ 2. List and explain 3 reasons why the...
First fill in your half cell and cell reactions. F in standard cell potentials as you calculate them. Oxidation at the Anode: The black (.) lead is attached to the electrode, which is the source of electrons. Write the anode half reaction: Reduction at the Cathode: The red lead (+) is attached to the electrode. Write the cathode half reaction: E degree - V. Overall Cell Reaction (Net Ionic equation): Write the overall cell reaction (balance electrons and add together):...
Part C Using standard potentials given in the appendices, calculate the standard cell potentials for the following reaction Zn(s) +Fe2(a)Zn2+ Fe(s) Report the answer in volts with three significant figures. LA Value Units Request Answer Submit Part D Determine the equilibrium constant for the above reaction Report to 1 significant figure and using scientific notation. ΑΣ Submit Request Answer
4(d) Explain briefly how standard half-cell reduction potentials are actually measured.
Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Ti2+]=0.00200 M and [Au3+]=0.753 M . Standard reduction potentials can be found in this table: https://sites.google.com/site/chempendix/potentials 3Ti(s)+2Au3+(aq)↽−−⇀3Ti2+(aq)+2Au(s)
Using standard potentials given in the appendices, calculate the standard cell potentials for the following reaction: Cu(s) +2Ag (aq)Cu2+ 2Ag(s) Report the answer in volts with three significant figures 0.442 Previous Answers Request Answer Submit X Incorrect; Try Again; 3 attempts remaining Review your calculations; you may have made a rounding error or used the wrong number of significant figures. Part B Determine the equilibrium constant for the above reaction Report to 1 significant figure and using scientific notation 9.1014
use Table 1 in the background section to
determine the 2 half-reactions and standard reduction potentials
for the redox reaction occurring in your galvanic cell. Record the
half reactions, identifying which is the oxidation and which is the
reduction half-reaction. Also record the corresponding reduction
potentials in Data Table 3.
Please help with Data Table 3
and
2. Was the electric potential found for your galvanic cell
consistent with the standard cell potential of the reaction (as
calculated in Data...
4. The following reactions are given, with their standard reduction potentials in V vs. NHE: CutteCu Cu2+ +2 eCu Explain why the sum of the first two reduction potentials doesn't equal the third. Consider the link between thermodynamic quantities (e.g. the Gibbs free energy) and cell potentials. 0.521 0.153 0.3419
Use the standard half-cell potentials listed below to calculate the standard cell potential and standard free energy Gº) for the following reaction occurring in an electrochemical cell at 25 °C. (The equation is balanced.) k(aq) + e-K(5) E* --2.93 V 12(s) + 2 0 - 2 (aq) E*-+0.54V a. +6.40 V & 670 KJ b. +1.85 V & 487K] C. 3.47 V & 241 Kb d. 3.47 V8 - 6709 e.+5,32 V &-+670K)
please help me answer this question........
thanks
Use the appropriate standard reduction potentials in the appendix of your book to determine the equilibrium constant at 205.00 K for the following reaction under acidic conditions. 4H^+(aq)+MnO_2(s)+2Fe^2+(aq) rightarrow Mn^2+(aq)+2Fe^3+(aq)+2H_2O(I) k = 6.19 times 10^53 The equation that relates equilibrium constants with cell potentials is: log k = nFE^0_cell/2.303RT Use this formula, along with the information in the appendix of your book to solve for K.