Why is the measured cell potential different from the theoretical potential?
There are many things which cause the cell potential to be less.Anything at all which causes any internal resistance in the cell or the circuit being used to measure it will reduce the measured potential.As well, frequently, non standard conditions may be used in a lab.The standard cells are for 1 M solutions. They are measured at 25 degrees Celsius. Differences in these conditions will cause changes. If a reactant or product is a gas, it needs to be bubbled over a platinum (inert) electrode with a pressure of 1 atm.Other much more significant differences also impact. Unless the electrodes are super clean, with a fresh metal surface that has no oxide layer at all, then the oxide layer will interfere with the reaction and reduce the measure potential. The salt bridge can cause interference to the easy flow of ions, and so cause internal resistance and lower the measured potential. How much will depend on a lot of factors about how it is made etc.If the connections within your circuit are not perfect, the measured potential will be less as well.I am sure you are beginning to imagine now that there are many factors which are very difficult to get just right.
What is the difference between oxidation potential and reduction potential?
oxidation potential:- potential or voltage required to take electrons and increase in their oxidation number is called oxidation potential and the reduction potential is just reverse of it.In the above picture they have given u the reduction potential and the oxidation potential will be just the negative of that value.
Can someone show me how to balance this chemical equation?
OK here is the easy way to deal with it,First work out how many electrons are needed to be transferred for the oxidant and the reducing agent, here we aim to make half equations2I- → I2 + 2e-IO3- + 6e- → something with I in the -1 oxidation stateNext adjust these equations to make them more reasonable in terms of converting one thing into something reasonable. The first one is OK, the second one needs protons and waters to be addedIO3- + 6e- + 6H+ → 3H2O + I-Now we combine the two half equations, to balance the number of electrons on both sidesIO3- + 6H+ 6e- + 6I- → 3I2 + 6e- + 3H2O + I-Now remove the electrons which appear on both sidesIO3- + 6H+ 6I- → 3I2 + 3H2O + I-Now remove one iodide from each side to give usIO3- + 6H+ 5I- → 3I2 + 3H2O
Chemistry half-reactions?
calculate if it's endothermic or not using H values (they're probably supplied somewhere), that will tell you if it's spontaneous. and look on a table of reduction potentials to see which is the stronger reducing agent, add that voltage to the negative of the weaker reducing agent... and that's the potential if it's at equilibrium.
Please oh please help me with these chemistry questions?
1. Consider the redox reaction: 3 Mn (s) + 2 Cr3+ (aq) → 3 Mn2+ (aq) + 2 Cr (s); if it has a ΔG° of -255 kJ, what is its E°? A. 0.881 V B. 4.404 × 10^-4 V C. 0.440 V D. 1.321 V 2. For a voltaic cell to deliver a potential, which must be correct? A. E° > 0, ΔG° < 0, K >1 B. E° > 0, ΔG° > 0, K <1 C. E° < 0, ΔG° > 0, K < 1 D. E° > 0, ΔG° < 0, K <1 3. Based on the standard electrode potentials shown below, which would be expected to be the strongest reducing agent? Half-reaction E° (V) Cu2+(aq) + 2e- → Cu(s) +0.337 Pb2+(aq) + 2e- → Pb(s) -0.126 Cd2+(aq) + 2e- → Cd(s) -0.403 Zn2+(aq) + 2e- → Zn(s) -0.763 A. Cu B. Cd C. Zn D. Pb 4. Which statement concerning batteries is true? A. A secondary cell can be recharged from an external power source after its voltage is dropped. B. The most common primary battery is the lithium-ion battery. C. A fuel cell is an example of a battery. D. When batteries are connected in a series, the total voltage is the product of their individual voltages. 5. Which statement about the anode in an electrolytic cell is correct? A. It has a positive polarity and is where oxidation occurs. B. It has a negative polarity and is where reduction occurs. C. It has a negative polarity and is where oxidation occurs. D. It has a positive polarity and is where reduction occurs.