1868 United States presidential election in Iowa

none June 2025 September 2023 1868 United States presidential election in Iowa presidential no 1864 United States presidential election in Iowa 1864 1872 United States presidential election in Iowa 1872 November 3, 1868 x200px Ulysses S Grant by Brady c1870-restored (3x4 crop).jpg Ulysses S. Grant Republican Party (United States) Illinois Schuyler Colfax 8 120,399 61.92% Hon. Horatio Seymour, N.Y - NARA - 528568 (cropped).jpg Horatio Seymour Democratic Party (United States) New York Francis Preston Blair Jr. 0 74,040 38.08% 1868 United States presidential election in Iowa results map by county.svg 312px County results Grant #e27f90 50–60% #cc2f4a 60–70% #d40000 70–80% #aa0000 80–90% #800000 90–100% Seymour #86b6f2 50–60% #4389e3 60–70% Tie #d4c4dc ~50% President Andrew Johnson Democratic Party (United States) Ulysses S. Grant Republican Party (United States) The 1868 United States presidential election in Iowa took place on November 3, 1868, as part of the 1868 United States presidential election. Voters chose eight representatives, or electors to the Electoral College, who voted for president and vice president. Iowa voted for the Republican nominee, Ulysses S. Grant, over the Democratic nominee, Horatio Seymour. Grant won the state by a margin of 23.84%. Results 1868 United States presidential election in Iowa1868 Presidential General Election Results - Iowa https://uselectionatlas.org/RESULTS/state.php?year=1868&fips=19&f=1&off=0&elect=0 Republican Party (United States) Ulysses S. Grant Illinois Schuyler Colfax Indiana 120,399 61.92% 8 100.00% Democratic Party (United States) Horatio Seymour New York Francis Preston Blair Jr. Missouri 74,040 38.08% 0 0.00% 194,439 100.00% 8 100.00% See also United States presidential elections in Iowa References Iowa 1868 Category:1868 Iowa elections