1864 United States presidential election in Rhode Island

none June 2025 September 2023 1864 United States presidential election 1864 United States presidential election in Rhode Island Rhode Island presidential no 1860 United States presidential election in Rhode Island 1860 1868 United States presidential election in Rhode Island 1868 November 8, 1864 x200px Abraham Lincoln O-84 by Brady, 1864 (Cropped).jpg Abraham Lincoln National Union Party (United States) Illinois Andrew Johnson 4 13,962 62.24% GeorgeMcClellan2 (3x4 cropped).jpg George B. McClellan Democratic Party (United States) New Jersey George H. Pendleton 0 8,470 37.76% 1864 United States presidential election in Rhode Island results map by county.svg 250px County results Lincoln #cc2f4a 60–70% President Abraham Lincoln Republican Party (United States) Abraham Lincoln National Union Party (United States) The 1864 United States presidential election in Rhode Island took place on November 8, 1864, as part of the 1864 United States presidential election. Voters chose four representatives, or electors to the Electoral College, who voted for president and vice president. Rhode Island voted for the National Union candidate, incumbent Republican Party President Abraham Lincoln and his running mate Andrew Johnson. They defeated the Democratic candidate, George B. McClellan and his running mate George H. Pendleton. Lincoln won the state by a margin of 24.48%. Results 1864 United States presidential election in Rhode Island1864 Presidential General Election Results - Rhode Island http://uselectionatlas.org/RESULTS/state.php?year=1864&fips=44&f=1&off=0&elect=0PartyCandidateVotesPercentageElectoral votes National Union Abraham Lincoln (incumbent) 13,962 62.24% 4 Democratic George B. McClellan 8,470 37.76% 0 Totals 22,432 100.0% 4 See also United States presidential elections in Rhode Island References Rhode Island 1864 Category:1864 Rhode Island elections