An MnO2(s) /Mn2+(aq) electrode in which the pH is 10.21 is prepared.
Part A: Find the [Mn2+] necessary to lower the potential of the half-cell to 0.00 V (at 25 ∘C).
An MnO2(s) /Mn2+(aq) electrode in which the pH is 10.21 is prepared. Part A: Find the...
An MnO2(s) /Mn2+(aq) electrode in which the pH is 10.27 is prepared.Find the [Mn2+] necessary to lower the potential of the half-cell to 0.00 V (at 25 ∘C).
Part A Consider a voltaic cell that is set up as follows: Anode contains an Zn(s) electrode and 1M Zn2+(aq) Cathode contains a Pt(s) electrode and 1M MnO4–(aq),Mn2+(aq), and H+(aq) Which of the following statements match the cathode? Select all that apply. Group of answer choices The electrode increases in mass Oxidation occurs at this half-cell Electrons enter the half-cell The electrode is inactive Cations from the salt-bridge move to this half-cell The electrode is negative Part B Which of...
Complete and balance the following half-reaction in basic solution Mn2+(aq) → MnO2(s)
If the potential of a hydrogen electrode based on the half-reaction 2 H^+ (aq) + 2 e^- rightarrow H_2 (g) is 0.000 V at pH = 0.00, what is the potential of the same electrode at pH = 6.75? Changing pH is changing the concentration of H^+. When the concentration is changed, the Nernst equation allows us to calculate the new cell potential.
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...
We consider two half cells: Half cell A comprises a platinum electrode immersed in an aqueous solution containing Mn2+ (10 mM) and MnO4- (4 mM) ions at pH=4. Half cell B is constructed with a silver electrode immersed in an aqueous potassium chromate solution (8 mM) in the presence of solid Ag2CrO4. The pH of this half cell B is pH=9. Both half cells are connected with a salt bridge. The voltage of the cell is measured at 25°C: it...
a.) A voltaic cell is constructed in which the cathode is a standard hydrogen electrode and the anode is a hydrogen electrode (P(H2) = 1 atm) immersed in a solution of unknown [H+]. If the cell potential is 0.204 V, what is the pH of the unknown solution at 298 K? b.) An electrochemical cell is constructed in which a Cr3+(1.00 M)|Cr(s) half-cell is connected to an H3O+(aq)|H2(1 atm) half-cell with unknown H3O+ concentration. The measured cell voltage is 0.366...
In the following cell, A is a standard Cu2+Cu electrode connected to a standard hydrogen electrode. If the voltmeter reading is +0.34 V, which half-reaction occurs in the left-hand cell compartment? Given: Standard reduction potential of the H1/H2 and Cu2*/Cu couples are 0.00 and +0.34 V, respectively. Holo H2(g) --> 2H+ (aq) + 2e7 2H(aq) + 2e --> H2(g) Cu(s) --> Cu2(aq) + 2e Cu2(aq) + 2e --> Cu(s)
A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)---- Sn(s)+Mn2+(aq) Calculate the cell potential of 25 degrees Celsius under each of the following conditions. Part A: Sn2+= 1.15*10^-2 M; and Mn2+= 2.37 M Part B: Sn2+= 2.37 M; and Mn2+= 1.15*10^-2
A voltaic cell employs the following redox reaction: Sn2+(aq)+Mn(s)→Sn(s)+Mn2+(aq) Calculate the cell potential at 25 ∘C∘C under each of the following conditions. Part A [Sn2+]= 1.34×10−2 MM ; [Mn2+]= 2.51 MM . Express your answer using two significant figures. Part B [Sn2+]=2.51 MM ; [Mn2+]=1.34×10−2 MM .